Post on 12-Sep-2015
MATH 38: Mathematical Analysis IIIUnit 3: Differentiation of Functions of More than One Variable
I. F. Evidente
IMSP (UPLB)
Outline
1 Relative Extreme Function Values
2 Absolute Extrema of Functions of Two Variables
Figures taken from: J. Stewart, The Calculus: Early Transcendentals,Brooks/Cole, 6th Edition, 2008.
Outline
1 Relative Extreme Function Values
2 Absolute Extrema of Functions of Two Variables
DefinitionGiven a function of two variables f (x, y) and a point (x0, y0) on thexy-plane:
1 f is said to have a relative maximum at (x0, y0) if
there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B .
2 f is said to have a relative minumum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B
3 f is said to have a relative extremum at (x0, y0) if either f has arelative maximum or a relative minimum at (x0, y0).
DefinitionGiven a function of two variables f (x, y) and a point (x0, y0) on thexy-plane:
1 f is said to have a relative maximum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such that
f (x0, y0) f (x, y) for every (x, y) B .2 f is said to have a relative minumum at (x0, y0) if there exists an
open ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B
3 f is said to have a relative extremum at (x0, y0) if either f has arelative maximum or a relative minimum at (x0, y0).
DefinitionGiven a function of two variables f (x, y) and a point (x0, y0) on thexy-plane:
1 f is said to have a relative maximum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B .
2 f is said to have a relative minumum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B
3 f is said to have a relative extremum at (x0, y0) if either f has arelative maximum or a relative minimum at (x0, y0).
DefinitionGiven a function of two variables f (x, y) and a point (x0, y0) on thexy-plane:
1 f is said to have a relative maximum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B .
2 f is said to have a relative minumum at (x0, y0) if
there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B
3 f is said to have a relative extremum at (x0, y0) if either f has arelative maximum or a relative minimum at (x0, y0).
DefinitionGiven a function of two variables f (x, y) and a point (x0, y0) on thexy-plane:
1 f is said to have a relative maximum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B .
2 f is said to have a relative minumum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such that
f (x0, y0) f (x, y) for every (x, y) B3 f is said to have a relative extremum at (x0, y0) if either f has a
relative maximum or a relative minimum at (x0, y0).
DefinitionGiven a function of two variables f (x, y) and a point (x0, y0) on thexy-plane:
1 f is said to have a relative maximum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B .
2 f is said to have a relative minumum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B
3 f is said to have a relative extremum at (x0, y0) if either f has arelative maximum or a relative minimum at (x0, y0).
DefinitionGiven a function of two variables f (x, y) and a point (x0, y0) on thexy-plane:
1 f is said to have a relative maximum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B .
2 f is said to have a relative minumum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B
3 f is said to have a relative extremum at (x0, y0) if
either f has arelative maximum or a relative minimum at (x0, y0).
DefinitionGiven a function of two variables f (x, y) and a point (x0, y0) on thexy-plane:
1 f is said to have a relative maximum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B .
2 f is said to have a relative minumum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B
3 f is said to have a relative extremum at (x0, y0) if either f has arelative maximum or a relative minimum at (x0, y0).
RemarkSuppose f has a relative minimum [maximum, extremum] at P .
The value f (x0, y0) is said to be a relative minimum [maximum,extremum] value of f .The point (x0, y0, f (x0, y0)) is called a relative minimum [maximum,extremum] point of f .
RemarkSuppose f has a relative minimum [maximum, extremum] at P .
The value f (x0, y0) is said to be a relative minimum [maximum,extremum] value of f .
The point (x0, y0, f (x0, y0)) is called a relative minimum [maximum,extremum] point of f .
RemarkSuppose f has a relative minimum [maximum, extremum] at P .
The value f (x0, y0) is said to be a relative minimum [maximum,extremum] value of f .The point (x0, y0, f (x0, y0)) is called a relative minimum [maximum,extremum] point of f .
Graphically, the graph of z = f (x, y) has peaks and trenches at relativemaximum and relative minimum points, respectively.
It is "relative" or "local" because the point is highest or lowest relative toits very close neighbors.Note that a function may have several relative extrema.
Graphically, the graph of z = f (x, y) has peaks and trenches at relativemaximum and relative minimum points, respectively.
It is "relative" or "local" because the point is highest or lowest relative toits very close neighbors.
Note that a function may have several relative extrema.
Graphically, the graph of z = f (x, y) has peaks and trenches at relativemaximum and relative minimum points, respectively.
It is "relative" or "local" because the point is highest or lowest relative toits very close neighbors.Note that a function may have several relative extrema.
Problem:How do we find the relative minima and relative maxima of f ?
DefinitionThe point (x0, y0) is a critical point of the function f (x, y) if
(x0, y0) dom f and either1 fx(x0, y0)= 0 and fy (x0, y0)= 0 (Type 1), or2 fx(x0, y0) or fy (x0, y0) does not exist (Type 2)
DefinitionThe point (x0, y0) is a critical point of the function f (x, y) if(x0, y0) dom f and either
1 fx(x0, y0)= 0 and fy (x0, y0)= 0 (Type 1), or2 fx(x0, y0) or fy (x0, y0) does not exist (Type 2)
DefinitionThe point (x0, y0) is a critical point of the function f (x, y) if(x0, y0) dom f and either
1 fx(x0, y0)= 0 and fy (x0, y0)= 0 (Type 1), or
2 fx(x0, y0) or fy (x0, y0) does not exist (Type 2)
DefinitionThe point (x0, y0) is a critical point of the function f (x, y) if(x0, y0) dom f and either
1 fx(x0, y0)= 0 and fy (x0, y0)= 0 (Type 1), or2 fx(x0, y0) or fy (x0, y0) does not exist (Type 2)
ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x2y and fy (x, y)= 2x+2y 8Critical Points:
6x2y = 02x+2y 8 = 0
} 4x8= 0 x = 2 & y = 6
(2,6) is a critical point of f .
ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x
2y and fy (x, y)= 2x+2y 8Critical Points:
6x2y = 02x+2y 8 = 0
} 4x8= 0 x = 2 & y = 6
(2,6) is a critical point of f .
ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x2y
and fy (x, y)= 2x+2y 8Critical Points:
6x2y = 02x+2y 8 = 0
} 4x8= 0 x = 2 & y = 6
(2,6) is a critical point of f .
ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x2y and fy (x, y)=
2x+2y 8Critical Points:
6x2y = 02x+2y 8 = 0
} 4x8= 0 x = 2 & y = 6
(2,6) is a critical point of f .
ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x2y and fy (x, y)= 2x
+2y 8Critical Points:
6x2y = 02x+2y 8 = 0
} 4x8= 0 x = 2 & y = 6
(2,6) is a critical point of f .
ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x2y and fy (x, y)= 2x+2y
8Critical Points:
6x2y = 02x+2y 8 = 0
} 4x8= 0 x = 2 & y = 6
(2,6) is a critical point of f .
ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x2y and fy (x, y)= 2x+2y 8
Critical Points:
6x2y = 02x+2y 8 = 0
} 4x8= 0 x = 2 & y = 6
(2,6) is a critical point of f .
ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x2y and fy (x, y)= 2x+2y 8Critical Points:
6x2y = 02x+2y 8 = 0
} 4x8= 0 x = 2 & y = 6
(2,6) is a critical point of f .
ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x2y and fy (x, y)= 2x+2y 8Critical Points:
6x2y = 02x+2y 8 = 0
}
4x8= 0 x = 2 & y = 6
(2,6) is a critical point of f .
ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x2y and fy (x, y)= 2x+2y 8Critical Points:
6x2y = 02x+2y 8 = 0
} 4x8= 0
x = 2 & y = 6
(2,6) is a critical point of f .
ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x2y and fy (x, y)= 2x+2y 8Critical Points:
6x2y = 02x+2y 8 = 0
} 4x8= 0 x = 2 &
y = 6
(2,6) is a critical point of f .
ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x2y and fy (x, y)= 2x+2y 8Critical Points:
6x2y = 02x+2y 8 = 0
} 4x8= 0 x = 2 & y = 6
(2,6) is a critical point of f .
ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x2y and fy (x, y)= 2x+2y 8Critical Points:
6x2y = 02x+2y 8 = 0
} 4x8= 0 x = 2 & y = 6
(2,6) is a critical point of f .
TheoremIf the function f (x, y) has a relative extremum at (x0,y0) and fx(x0, y0) andfy (x0, y0) exist, then
fx(x0, y0)= 0 and fy (x0, y0)= 0.
TheoremIf the function f (x, y) has a relative extremum at (x0,y0) and fx(x0, y0) andfy (x0, y0) exist, then fx(x0, y0)= 0 and fy (x0, y0)= 0.
The theorem tells us the following:
RemarkIf f (x, y) has a relative extrema at (x0, y0) and the partial derivativesof f exist at that point, then (x0, y0) must be a Type 1 critical point.If f is differentiable for all points in its domain, then the candidatesfor points where relative extrema occur come from the critical pointsof f . (Naturally, these are all Type 1 critical points) Venn Diagram?The converse of the theorem is NOT true. It is not true that if (x0, y0)is a critical point of f , then f has a relative extrema at (x0, y0).
The theorem tells us the following:
RemarkIf f (x, y) has a relative extrema at (x0, y0) and the partial derivativesof f exist at that point, then (x0, y0) must be
a Type 1 critical point.If f is differentiable for all points in its domain, then the candidatesfor points where relative extrema occur come from the critical pointsof f . (Naturally, these are all Type 1 critical points) Venn Diagram?The converse of the theorem is NOT true. It is not true that if (x0, y0)is a critical point of f , then f has a relative extrema at (x0, y0).
The theorem tells us the following:
RemarkIf f (x, y) has a relative extrema at (x0, y0) and the partial derivativesof f exist at that point, then (x0, y0) must be a Type 1 critical point.
If f is differentiable for all points in its domain, then the candidatesfor points where relative extrema occur come from the critical pointsof f . (Naturally, these are all Type 1 critical points) Venn Diagram?The converse of the theorem is NOT true. It is not true that if (x0, y0)is a critical point of f , then f has a relative extrema at (x0, y0).
The theorem tells us the following:
RemarkIf f (x, y) has a relative extrema at (x0, y0) and the partial derivativesof f exist at that point, then (x0, y0) must be a Type 1 critical point.If f is differentiable for all points in its domain, then the candidatesfor points where relative extrema occur come from
the critical pointsof f . (Naturally, these are all Type 1 critical points) Venn Diagram?The converse of the theorem is NOT true. It is not true that if (x0, y0)is a critical point of f , then f has a relative extrema at (x0, y0).
The theorem tells us the following:
RemarkIf f (x, y) has a relative extrema at (x0, y0) and the partial derivativesof f exist at that point, then (x0, y0) must be a Type 1 critical point.If f is differentiable for all points in its domain, then the candidatesfor points where relative extrema occur come from the critical pointsof f .
(Naturally, these are all Type 1 critical points) Venn Diagram?The converse of the theorem is NOT true. It is not true that if (x0, y0)is a critical point of f , then f has a relative extrema at (x0, y0).
The theorem tells us the following:
RemarkIf f (x, y) has a relative extrema at (x0, y0) and the partial derivativesof f exist at that point, then (x0, y0) must be a Type 1 critical point.If f is differentiable for all points in its domain, then the candidatesfor points where relative extrema occur come from the critical pointsof f . (Naturally, these are all Type 1 critical points)
Venn Diagram?The converse of the theorem is NOT true. It is not true that if (x0, y0)is a critical point of f , then f has a relative extrema at (x0, y0).
The theorem tells us the following:
RemarkIf f (x, y) has a relative extrema at (x0, y0) and the partial derivativesof f exist at that point, then (x0, y0) must be a Type 1 critical point.If f is differentiable for all points in its domain, then the candidatesfor points where relative extrema occur come from the critical pointsof f . (Naturally, these are all Type 1 critical points) Venn Diagram?
The converse of the theorem is NOT true. It is not true that if (x0, y0)is a critical point of f , then f has a relative extrema at (x0, y0).
The theorem tells us the following:
RemarkIf f (x, y) has a relative extrema at (x0, y0) and the partial derivativesof f exist at that point, then (x0, y0) must be a Type 1 critical point.If f is differentiable for all points in its domain, then the candidatesfor points where relative extrema occur come from the critical pointsof f . (Naturally, these are all Type 1 critical points) Venn Diagram?The converse of the theorem is NOT true. It is not true that if (x0, y0)is a critical point of f , then f has a relative extrema at (x0, y0).
Geometrically
Suppose the following:z = f (x, y) has a relative extremum at (x0, y0)The tangent plane of the graph of f at this point exists.
fx(x0, y0)= 0 and fy (x0, y0)= 0Equation of tangent plane:
z = L(x, y)z = fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)z = f (x0, y0)
What kind of plane is this? HORIZONTAL!
RemarkIf the tangent plane of the graph of f at a relative extreme point exists,then it must be horizontal!
Geometrically
Suppose the following:z = f (x, y) has a relative extremum at (x0, y0)The tangent plane of the graph of f at this point exists.
fx(x0, y0)= 0 and fy (x0, y0)= 0
Equation of tangent plane:
z = L(x, y)z = fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)z = f (x0, y0)
What kind of plane is this? HORIZONTAL!
RemarkIf the tangent plane of the graph of f at a relative extreme point exists,then it must be horizontal!
Geometrically
Suppose the following:z = f (x, y) has a relative extremum at (x0, y0)The tangent plane of the graph of f at this point exists.
fx(x0, y0)= 0 and fy (x0, y0)= 0Equation of tangent plane:
z = L(x, y)
z = fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)z = f (x0, y0)
What kind of plane is this? HORIZONTAL!
RemarkIf the tangent plane of the graph of f at a relative extreme point exists,then it must be horizontal!
Geometrically
Suppose the following:z = f (x, y) has a relative extremum at (x0, y0)The tangent plane of the graph of f at this point exists.
fx(x0, y0)= 0 and fy (x0, y0)= 0Equation of tangent plane:
z = L(x, y)z = fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)
z = f (x0, y0)
What kind of plane is this? HORIZONTAL!
RemarkIf the tangent plane of the graph of f at a relative extreme point exists,then it must be horizontal!
Geometrically
Suppose the following:z = f (x, y) has a relative extremum at (x0, y0)The tangent plane of the graph of f at this point exists.
fx(x0, y0)= 0 and fy (x0, y0)= 0Equation of tangent plane:
z = L(x, y)z = fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)z = f (x0, y0)
What kind of plane is this? HORIZONTAL!
RemarkIf the tangent plane of the graph of f at a relative extreme point exists,then it must be horizontal!
Geometrically
Suppose the following:z = f (x, y) has a relative extremum at (x0, y0)The tangent plane of the graph of f at this point exists.
fx(x0, y0)= 0 and fy (x0, y0)= 0Equation of tangent plane:
z = L(x, y)z = fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)z = f (x0, y0)
What kind of plane is this?
HORIZONTAL!
RemarkIf the tangent plane of the graph of f at a relative extreme point exists,then it must be horizontal!
Geometrically
Suppose the following:z = f (x, y) has a relative extremum at (x0, y0)The tangent plane of the graph of f at this point exists.
fx(x0, y0)= 0 and fy (x0, y0)= 0Equation of tangent plane:
z = L(x, y)z = fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)z = f (x0, y0)
What kind of plane is this? HORIZONTAL!
RemarkIf the tangent plane of the graph of f at a relative extreme point exists,then it must be horizontal!
Geometrically
Suppose the following:z = f (x, y) has a relative extremum at (x0, y0)The tangent plane of the graph of f at this point exists.
fx(x0, y0)= 0 and fy (x0, y0)= 0Equation of tangent plane:
z = L(x, y)z = fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)z = f (x0, y0)
What kind of plane is this? HORIZONTAL!
RemarkIf the tangent plane of the graph of f at a relative extreme point exists,then
it must be horizontal!
Geometrically
Suppose the following:z = f (x, y) has a relative extremum at (x0, y0)The tangent plane of the graph of f at this point exists.
fx(x0, y0)= 0 and fy (x0, y0)= 0Equation of tangent plane:
z = L(x, y)z = fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)z = f (x0, y0)
What kind of plane is this? HORIZONTAL!
RemarkIf the tangent plane of the graph of f at a relative extreme point exists,then it must be horizontal!
DefinitionA point (x0, y0,z0) is said to be a saddle point of a surface if
there existtwo vertical planes such that the trace of in one plane has a relativemaximum at (x0, y0) and the trace of in the other plane has a relativeminimum at (x0, y0).
DefinitionA point (x0, y0,z0) is said to be a saddle point of a surface if there existtwo vertical planes such that
the trace of in one plane has a relativemaximum at (x0, y0) and the trace of in the other plane has a relativeminimum at (x0, y0).
DefinitionA point (x0, y0,z0) is said to be a saddle point of a surface if there existtwo vertical planes such that the trace of in one plane has a relativemaximum at (x0, y0) and
the trace of in the other plane has a relativeminimum at (x0, y0).
DefinitionA point (x0, y0,z0) is said to be a saddle point of a surface if there existtwo vertical planes such that the trace of in one plane has a relativemaximum at (x0, y0) and the trace of in the other plane has a relativeminimum at (x0, y0).
f (x, y)= x2 y2 has a saddle point at (0,0)
Theorem (Second Derivative Test for Relative Extrema)Suppose that the first and second order partial derivatives of adifferentiable function f (x, y) is continuous on some open ball B
((x0, y0),r
)and fx(x0, y0)= fy (x0, y0)= 0.
Let
D = fxx(x0, y0) fy y (x0, y0)[fxy (x0, y0)
]2 .1 If D > 0 and
1 fxx (x0, y0)> 0, then f (x, y) has a relative minimum at (x0, y0).2 fxx (x0, y0)< 0, then f (x, y) has a relative maximum at (x0, y0).
2 If D < 0, then f (x, y) has a saddle point at (x0, y0).3 If D = 0, then no conclusion can be made.
Theorem (Second Derivative Test for Relative Extrema)Suppose that the first and second order partial derivatives of adifferentiable function f (x, y) is continuous on some open ball B
((x0, y0),r
)and fx(x0, y0)= fy (x0, y0)= 0. Let
D = fxx(x0, y0) fy y (x0, y0)[fxy (x0, y0)
]2 .
1 If D > 0 and1 fxx (x0, y0)> 0, then f (x, y) has a relative minimum at (x0, y0).2 fxx (x0, y0)< 0, then f (x, y) has a relative maximum at (x0, y0).
2 If D < 0, then f (x, y) has a saddle point at (x0, y0).3 If D = 0, then no conclusion can be made.
Theorem (Second Derivative Test for Relative Extrema)Suppose that the first and second order partial derivatives of adifferentiable function f (x, y) is continuous on some open ball B
((x0, y0),r
)and fx(x0, y0)= fy (x0, y0)= 0. Let
D = fxx(x0, y0) fy y (x0, y0)[fxy (x0, y0)
]2 .1 If D > 0 and
1 fxx (x0, y0)> 0, then f (x, y) has a relative minimum at (x0, y0).2 fxx (x0, y0)< 0, then f (x, y) has a relative maximum at (x0, y0).
2 If D < 0, then f (x, y) has a saddle point at (x0, y0).3 If D = 0, then no conclusion can be made.
Theorem (Second Derivative Test for Relative Extrema)Suppose that the first and second order partial derivatives of adifferentiable function f (x, y) is continuous on some open ball B
((x0, y0),r
)and fx(x0, y0)= fy (x0, y0)= 0. Let
D = fxx(x0, y0) fy y (x0, y0)[fxy (x0, y0)
]2 .1 If D > 0 and
1 fxx (x0, y0)> 0, then
f (x, y) has a relative minimum at (x0, y0).2 fxx (x0, y0)< 0, then f (x, y) has a relative maximum at (x0, y0).
2 If D < 0, then f (x, y) has a saddle point at (x0, y0).3 If D = 0, then no conclusion can be made.
Theorem (Second Derivative Test for Relative Extrema)Suppose that the first and second order partial derivatives of adifferentiable function f (x, y) is continuous on some open ball B
((x0, y0),r
)and fx(x0, y0)= fy (x0, y0)= 0. Let
D = fxx(x0, y0) fy y (x0, y0)[fxy (x0, y0)
]2 .1 If D > 0 and
1 fxx (x0, y0)> 0, then f (x, y) has a relative minimum at (x0, y0).
2 fxx (x0, y0)< 0, then f (x, y) has a relative maximum at (x0, y0).2 If D < 0, then f (x, y) has a saddle point at (x0, y0).3 If D = 0, then no conclusion can be made.
Theorem (Second Derivative Test for Relative Extrema)Suppose that the first and second order partial derivatives of adifferentiable function f (x, y) is continuous on some open ball B
((x0, y0),r
)and fx(x0, y0)= fy (x0, y0)= 0. Let
D = fxx(x0, y0) fy y (x0, y0)[fxy (x0, y0)
]2 .1 If D > 0 and
1 fxx (x0, y0)> 0, then f (x, y) has a relative minimum at (x0, y0).2 fxx (x0, y0)< 0, then
f (x, y) has a relative maximum at (x0, y0).2 If D < 0, then f (x, y) has a saddle point at (x0, y0).3 If D = 0, then no conclusion can be made.
Theorem (Second Derivative Test for Relative Extrema)Suppose that the first and second order partial derivatives of adifferentiable function f (x, y) is continuous on some open ball B
((x0, y0),r
)and fx(x0, y0)= fy (x0, y0)= 0. Let
D = fxx(x0, y0) fy y (x0, y0)[fxy (x0, y0)
]2 .1 If D > 0 and
1 fxx (x0, y0)> 0, then f (x, y) has a relative minimum at (x0, y0).2 fxx (x0, y0)< 0, then f (x, y) has a relative maximum at (x0, y0).
2 If D < 0, then f (x, y) has a saddle point at (x0, y0).3 If D = 0, then no conclusion can be made.
Theorem (Second Derivative Test for Relative Extrema)Suppose that the first and second order partial derivatives of adifferentiable function f (x, y) is continuous on some open ball B
((x0, y0),r
)and fx(x0, y0)= fy (x0, y0)= 0. Let
D = fxx(x0, y0) fy y (x0, y0)[fxy (x0, y0)
]2 .1 If D > 0 and
1 fxx (x0, y0)> 0, then f (x, y) has a relative minimum at (x0, y0).2 fxx (x0, y0)< 0, then f (x, y) has a relative maximum at (x0, y0).
2 If D < 0, then
f (x, y) has a saddle point at (x0, y0).3 If D = 0, then no conclusion can be made.
Theorem (Second Derivative Test for Relative Extrema)Suppose that the first and second order partial derivatives of adifferentiable function f (x, y) is continuous on some open ball B
((x0, y0),r
)and fx(x0, y0)= fy (x0, y0)= 0. Let
D = fxx(x0, y0) fy y (x0, y0)[fxy (x0, y0)
]2 .1 If D > 0 and
1 fxx (x0, y0)> 0, then f (x, y) has a relative minimum at (x0, y0).2 fxx (x0, y0)< 0, then f (x, y) has a relative maximum at (x0, y0).
2 If D < 0, then f (x, y) has a saddle point at (x0, y0).
3 If D = 0, then no conclusion can be made.
Theorem (Second Derivative Test for Relative Extrema)Suppose that the first and second order partial derivatives of adifferentiable function f (x, y) is continuous on some open ball B
((x0, y0),r
)and fx(x0, y0)= fy (x0, y0)= 0. Let
D = fxx(x0, y0) fy y (x0, y0)[fxy (x0, y0)
]2 .1 If D > 0 and
1 fxx (x0, y0)> 0, then f (x, y) has a relative minimum at (x0, y0).2 fxx (x0, y0)< 0, then f (x, y) has a relative maximum at (x0, y0).
2 If D < 0, then f (x, y) has a saddle point at (x0, y0).3 If D = 0, then
no conclusion can be made.
Theorem (Second Derivative Test for Relative Extrema)Suppose that the first and second order partial derivatives of adifferentiable function f (x, y) is continuous on some open ball B
((x0, y0),r
)and fx(x0, y0)= fy (x0, y0)= 0. Let
D = fxx(x0, y0) fy y (x0, y0)[fxy (x0, y0)
]2 .1 If D > 0 and
1 fxx (x0, y0)> 0, then f (x, y) has a relative minimum at (x0, y0).2 fxx (x0, y0)< 0, then f (x, y) has a relative maximum at (x0, y0).
2 If D < 0, then f (x, y) has a saddle point at (x0, y0).3 If D = 0, then no conclusion can be made.
RemarkThis test is only for Type 1 critical points!If f is differentiable, then this test allows you to get ALL criticalpoints.
RemarkOne way to remember the formula for D:
The Jacobian of f is the matrix of partial derivatives of f[fxx fxyfyx fy y
]D is the determinant of the Jacobian of f .
TheoremLet f be a function in 2 variables and R a region on the xy-plane. If fxyand fyx are continuous on R, then fxy = fyx on that open ball.
RemarkOne way to remember the formula for D:The Jacobian of f is the matrix of partial derivatives of f[
fxx fxyfyx fy y
]
D is the determinant of the Jacobian of f .
TheoremLet f be a function in 2 variables and R a region on the xy-plane. If fxyand fyx are continuous on R, then fxy = fyx on that open ball.
RemarkOne way to remember the formula for D:The Jacobian of f is the matrix of partial derivatives of f[
fxx fxyfyx fy y
]D is the determinant of the Jacobian of f .
TheoremLet f be a function in 2 variables and R a region on the xy-plane. If fxyand fyx are continuous on R, then fxy = fyx on that open ball.
RemarkOne way to remember the formula for D:The Jacobian of f is the matrix of partial derivatives of f[
fxx fxyfyx fy y
]D is the determinant of the Jacobian of f .
TheoremLet f be a function in 2 variables and R a region on the xy-plane. If fxyand fyx are continuous on R, then fxy = fyx on that open ball.
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8
Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point:
(2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)
Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)=
6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6
fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)=
2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2
fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)=
2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2
Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6
22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6
22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 2
2 2= 124= 8> 0 (rel ext)
Nowfxx(2,6)= 6> 0
f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22
2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
=
124= 8> 0 (rel ext)
Nowfxx(2,6)= 6> 0
f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 12
4= 8> 0 (rel ext)
Nowfxx(2,6)= 6> 0
f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124=
8> 0 (rel ext)
Nowfxx(2,6)= 6> 0
f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8
> 0 (rel ext)
Nowfxx(2,6)= 6> 0
f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0
(rel ext)
Nowfxx(2,6)= 6> 0
f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)
Nowfxx(2,6)= 6> 0
f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)=
6> 0 f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6
> 0 f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0
f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
ExampleLocate all relative extrema and saddle points of f (x, y)= 4xy x4 y4.
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points:
4y 4x3 = 0 (1)4x4y3 = 0 (2)
} y = x
3 (1)x = y3 (2)
}
ExampleLocate all relative extrema and saddle points of f (x, y)= 4xy x4 y4.
fx(x, y)=
4y 4x3 and fy (x, y)= 4x4y3
Critical Points:
4y 4x3 = 0 (1)4x4y3 = 0 (2)
} y = x
3 (1)x = y3 (2)
}
ExampleLocate all relative extrema and saddle points of f (x, y)= 4xy x4 y4.
fx(x, y)= 4y 4x3
and fy (x, y)= 4x4y3
Critical Points:
4y 4x3 = 0 (1)4x4y3 = 0 (2)
} y = x
3 (1)x = y3 (2)
}
ExampleLocate all relative extrema and saddle points of f (x, y)= 4xy x4 y4.
fx(x, y)= 4y 4x3 and
fy (x, y)= 4x4y3
Critical Points:
4y 4x3 = 0 (1)4x4y3 = 0 (2)
} y = x
3 (1)x = y3 (2)
}
ExampleLocate all relative extrema and saddle points of f (x, y)= 4xy x4 y4.
fx(x, y)= 4y 4x3 and fy (x, y)=
4x4y3
Critical Points:
4y 4x3 = 0 (1)4x4y3 = 0 (2)
} y = x
3 (1)x = y3 (2)
}
ExampleLocate all relative extrema and saddle points of f (x, y)= 4xy x4 y4.
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points:
4y 4x3 = 0 (1)4x4y3 = 0 (2)
} y = x
3 (1)x = y3 (2)
}
ExampleLocate all relative extrema and saddle points of f (x, y)= 4xy x4 y4.
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points:
4y 4x3 = 0 (1)4x4y3 = 0 (2)
} y = x
3 (1)x = y3 (2)
}
ExampleLocate all relative extrema and saddle points of f (x, y)= 4xy x4 y4.
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points:
4y 4x3 = 0 (1)4x4y3 = 0 (2)
}
y = x3 (1)
x = y3 (2)}
ExampleLocate all relative extrema and saddle points of f (x, y)= 4xy x4 y4.
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points:
4y 4x3 = 0 (1)4x4y3 = 0 (2)
} y = x
3 (1)x = y3 (2)
}
y = x3 (1)x = y3 (2)
}
Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0 y =1 y = 1
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0 y =1 y = 1
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3
x = x9x9x = 0
x(x81) = 0x(x41)(x4+1) = 0
x(x21)(x2+1)(x4+1) = 0x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0 y =1 y = 1
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0 y =1 y = 1
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0
x(x81) = 0x(x41)(x4+1) = 0
x(x21)(x2+1)(x4+1) = 0x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0 y =1 y = 1
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0 y =1 y = 1
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0
x(x21)(x2+1)(x4+1) = 0x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0 y =1 y = 1
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0 y =1 y = 1
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0 y =1 y = 1
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0 y =1 y = 1
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0
x =1 x = 1y = 0 y =1 y = 1
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1
x = 1y = 0 y =1 y = 1
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1
y = 0 y =1 y = 1
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0
y =1 y = 1
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0 y =1
y = 1
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0 y =1 y = 1
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2
0 12 12
fy y (x, y)=12y2
0 12 12
fxy (x, y)= 4
4 4 4
D
16 128 128
conclusion
saddle pt rel max rel max
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0
12 12
fy y (x, y)=12y2
0 12 12
fxy (x, y)= 4
4 4 4
D
16 128 128
conclusion
saddle pt rel max rel max
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12
12
fy y (x, y)=12y2
0 12 12
fxy (x, y)= 4
4 4 4
D
16 128 128
conclusion
saddle pt rel max rel max
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2
0 12 12
fxy (x, y)= 4
4 4 4
D
16 128 128
conclusion
saddle pt rel max rel max
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0
12 12
fxy (x, y)= 4
4 4 4
D
16 128 128
conclusion
saddle pt rel max rel max
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12
12
fxy (x, y)= 4
4 4 4
D
16 128 128
conclusion
saddle pt rel max rel max
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12
fxy (x, y)= 4
4 4 4
D
16 128 128
conclusion
saddle pt rel max rel max
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12
fxy (x, y)= 4 4
4 4
D
16 128 128
conclusion
saddle pt rel max rel max
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12
fxy (x, y)= 4 4 4
4
D
16 128 128
conclusion
saddle pt rel max rel max
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12
fxy (x, y)= 4 4 4 4D
16 128 128
conclusion
saddle pt rel max rel max
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12
fxy (x, y)= 4 4 4 4D 16
128 128
conclusion
saddle pt rel max rel max
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12
fxy (x, y)= 4 4 4 4D 16 128
128
conclusion
saddle pt rel max rel max
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12
fxy (x, y)= 4 4 4 4D 16 128 128
conclusion
saddle pt rel max rel max
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12
fxy (x, y)= 4 4 4 4D 16 128 128
conclusion saddle pt
rel max rel max
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12
fxy (x, y)= 4 4 4 4D 16 128 128
conclusion saddle pt rel max
rel max
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12
fxy (x, y)= 4 4 4 4D 16 128 128
conclusion saddle pt rel max rel max
Outline
1 Relative Extreme Function Values
2 Absolute Extrema of Functions of Two Variables
DefinitionLet f be a function of two variables x and y . Let R be a region on thexy-plane and (x0, y0) a point in R.
1 f is said to have an absolute maximum on R at (x0, y0) iff (x0, y0) f (x, y) for every (x, y) R.
2 f is said to have an absolute minimum on R at (x0, y0) iff (x0, y0) f (x, y) for every (x, y) R.
3 f is said to have an absolute extremum on R at (x0, y0) if either fhas an absolute maximum value or absolute minimum value on R at(x0, y0).
DefinitionLet f be a function of two variables x and y . Let R be a region on thexy-plane and (x0, y0) a point in R.
1 f is said to have an absolute maximum on R at (x0, y0) iff (x0, y0) f (x, y) for every (x, y) R.
2 f is said to have an absolute minimum on R at (x0, y0) iff (x0, y0) f (x, y) for every (x, y) R.
3 f is said to have an absolute extremum on R at (x0, y0) if either fhas an absolute maximum value or absolute minimum value on R at(x0, y0).
DefinitionLet f be a function of two variables x and y . Let R be a region on thexy-plane and (x0, y0) a point in R.
1 f is said to have an absolute maximum on R at (x0, y0) iff (x0, y0) f (x, y) for every (x, y) R.
2 f is said to have an absolute minimum on R at (x0, y0) iff (x0, y0) f (x, y) for every (x, y) R.
3 f is said to have an absolute extremum on R at (x0, y0) if either fhas an absolute maximum value or absolute minimum value on R at(x0, y0).
DefinitionLet f be a function of two variables x and y . Let R be a region on thexy-plane and (x0, y0) a point in R.
1 f is said to have an absolute maximum on R at (x0, y0) iff (x0, y0) f (x, y) for every (x, y) R.
2 f is said to have an absolute minimum on R at (x0, y0) iff (x0, y0) f (x, y) for every (x, y) R.
3 f is said to have an absolute extremum on R at (x0, y0) if either fhas an absolute maximum value or absolute minimum value on R at(x0, y0).
DefinitionLet f be a function of two variables x and y . Let R be a region on thexy-plane and (x0, y0) a point in R.
1 f is said to have an absolute maximum on R at (x0, y0) iff (x0, y0) f (x, y) for every (x, y) R.
2 f is said to have an absolute minimum on R at (x0, y0) iff (x0, y0) f (x, y) for every (x, y) R.
3 f is said to have an absolute extremum on R at (x0, y0) if either fhas an absolute maximum value or absolute minimum value on R at(x0, y0).
RemarkSuppose that f has an absolute maximum [minimum, extremum] at(x0, y0).
1 We say f (x0, y0) is the absolute maximum [minimum, extremum] valueof f on R.
2 What is the difference between a relative extremum point and anabsolute extremum point?
RemarkSuppose that f has an absolute maximum [minimum, extremum] at(x0, y0).
1 We say f (x0, y0) is the absolute maximum [minimum, extremum] valueof f on R.
2 What is the difference between a relative extremum point and anabsolute extremum point?
RemarkSuppose that f has an absolute maximum [minimum, extremum] at(x0, y0).
1 We say f (x0, y0) is the absolute maximum [minimum, extremum] valueof f on R.
2 What is the difference between a relative extremum point and anabsolute extremum point?
RemarkSuppose that f has an absolute maximum [minimum, extremum] at(x0, y0).
1 We say f (x0, y0) is the absolute maximum [minimum, extremum] valueof f on R.
2 What is the difference between a relative extremum point and anabsolute extremum point?
DefinitionLet R be a region in the xy-plane
1 R is bounded if it can be contained in some rectangle.2 R closed if R contains its boundary.
DefinitionLet R be a region in the xy-plane
1 R is bounded if it can be contained in some rectangle.
2 R closed if R contains its boundary.
DefinitionLet R be a region in the xy-plane
1 R is bounded if it can be contained in some rectangle.2 R closed if R contains its boundary.
Theorem (Extreme Value Theorem)If f (x, y) is continuous on a closed and bounded set R, then
f has both anabsolute maximum and an absolute minimum on R.
Theorem (Extreme Value Theorem)If f (x, y) is continuous on a closed and bounded set R, then f has both anabsolute maximum and an absolute minimum on R.
RemarkIf f and R satisfy the conditions of EVT, the absolute extrema occur at
1 critical points inside R
2 points on the boundary of R
RemarkIf f and R satisfy the conditions of EVT, the absolute extrema occur at
1 critical points inside R2 points on the boundary of R
ProcedureFinding Absolute Extrema on a Region R satisfying EVT:
1 Find the critical numbers of f (x, y) in R.
Evaluate f (x, y) at thecritical numbers.
2 Evaluate f (x, y) at the vertices of R (if any).3 Convert f (x, y) to a function of one variable g (x) or g
(y)on the
boundaries of R.4 Find the critical numbers of g and evaluate f (x, y) at these numbers.
Recall: For a function of one variable, the critical numbers are pointsfor which the derivative of g is 0 or does not exist.
5 Compare the values obtained.
ProcedureFinding Absolute Extrema on a Region R satisfying EVT:
1 Find the critical numbers of f (x, y) in R. Evaluate f (x, y) at thecritical numbers.
2 Evaluate f (x, y) at the vertices of R (if any).3 Convert f (x, y) to a function of one variable g (x) or g
(y)on the
boundaries of R.4 Find the critical numbers of g and evaluate f (x, y) at these numbers.
Recall: For a function of one variable, the critical numbers are pointsfor which the derivative of g is 0 or does not exist.
5 Compare the values obtained.
ProcedureFinding Absolute Extrema on a Region R satisfying EVT:
1 Find the critical numbers of f (x, y) in R. Evaluate f (x, y) at thecritical numbers.
2 Evaluate f (x, y) at the vertices of R (if any).
3 Convert f (x, y) to a function of one variable g (x) or g(y)on the
boundaries of R.4 Find the critical numbers of g and evaluate f (x, y) at these numbers.
Recall: For a function of one variable, the critical numbers are pointsfor which the derivative of g is 0 or does not exist.
5 Compare the values obtained.
ProcedureFinding Absolute Extrema on a Region R satisfying EVT:
1 Find the critical numbers of f (x, y) in R. Evaluate f (x, y) at thecritical numbers.
2 Evaluate f (x, y) at the vertices of R (if any).3 Convert f (x, y) to a function of one variable g (x) or g
(y)on the
boundaries of R.
4 Find the critical numbers of g and evaluate f (x, y) at these numbers.Recall: For a function of one variable, the critical numbers are pointsfor which the derivative of g is 0 or does not exist.
5 Compare the values obtained.
ProcedureFinding Absolute Extrema on a Region R satisfying EVT:
1 Find the critical numbers of f (x, y) in R. Evaluate f (x, y) at thecritical numbers.
2 Evaluate f (x, y) at the vertices of R (if any).3 Convert f (x, y) to a function of one variable g (x) or g
(y)on the
boundaries of R.4 Find the critical numbers of g and evaluate f (x, y) at these numbers.
Recall: For a function of one variable, the critical numbers are pointsfor which the derivative of g is 0 or does not exist.
5 Compare the values obtained.
ProcedureFinding Absolute Extrema on a Region R satisfying EVT:
1 Find the critical numbers of f (x, y) in R. Evaluate f (x, y) at thecritical numbers.
2 Evaluate f (x, y) at the vertices of R (if any).3 Convert f (x, y) to a function of one variable g (x) or g
(y)on the
boundaries of R.4 Find the critical numbers of g and evaluate f (x, y) at these numbers.
Recall: For a function of one variable, the critical numbers are pointsfor which the derivative of g is 0 or does not exist.
5 Compare the values obtained.
ProcedureFinding Absolute Extrema on a Region R satisfying EVT:
1 Find the critical numbers of f (x, y) in R. Evaluate f (x, y) at thecritical numbers.
2 Evaluate f (x, y) at the vertices of R (if any).3 Convert f (x, y) to a function of one variable g (x) or g
(y)on the
boundaries of R.4 Find the critical numbers of g and evaluate f (x, y) at these numbers.
Recall: For a function of one variable, the critical numbers are pointsfor which the derivative of g is 0 or does not exist.
5 Compare the values obtained.
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)= 2x1 and fy (x, y)= 4yCritical Point:
(12 ,0
)At critical point inside R: f
(12 ,0
)= 14At vertices: noneAt boundary: f
(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)=
2x1 and fy (x, y)= 4yCritical Point:
(12 ,0
)At critical point inside R: f
(12 ,0
)= 14At vertices: noneAt boundary: f
(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)= 2x1
and fy (x, y)= 4yCritical Point:
(12 ,0
)At critical point inside R: f
(12 ,0
)= 14At vertices: noneAt boundary: f
(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)= 2x1 and
fy (x, y)= 4yCritical Point:
(12 ,0
)At critical point inside R: f
(12 ,0
)= 14At vertices: noneAt boundary: f
(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)= 2x1 and fy (x, y)=
4y
Critical Point:(12 ,0
)At critical point inside R: f
(12 ,0
)= 14At vertices: noneAt boundary: f
(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)= 2x1 and fy (x, y)= 4y
Critical Point:(12 ,0
)At critical point inside R: f
(12 ,0
)= 14At vertices: noneAt boundary: f
(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)= 2x1 and fy (x, y)= 4yCritical Point:
(12 ,0
)
At critical point inside R: f(12 ,0
)= 14At vertices: noneAt boundary: f
(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)= 2x1 and fy (x, y)= 4yCritical Point:
(12 ,0
)At critical point inside R: f
(12 ,0
)=
14At vertices: noneAt boundary: f
(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)= 2x1 and fy (x, y)= 4yCritical Point:
(12 ,0
)At critical point inside R: f
(12 ,0
)= 14
At vertices: noneAt boundary: f
(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)= 2x1 and fy (x, y)= 4yCritical Point:
(12 ,0
)At critical point inside R: f
(12 ,0
)= 14At vertices:
noneAt boundary: f
(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)= 2x1 and fy (x, y)= 4yCritical Point:
(12 ,0
)At critical point inside R: f
(12 ,0
)= 14At vertices: none
At boundary: f(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)= 2x1 and fy (x, y)= 4yCritical Point:
(12 ,0
)At critical point inside R: f
(12 ,0
)= 14At vertices: noneAt boundary: f
(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)= 2x1 and fy (x, y)= 4yCritical Point:
(12 ,0
)At critical point inside R: f
(12 ,0
)= 14At vertices: noneAt boundary: f
(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)=
3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6
and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and
fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)=
3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)= 3x3
Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)
At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)=
1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1
At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:
f (0,0)= 7 f (0,5)= 8 f (3,0)= 11
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)=
7 f (0,5)= 8 f (3,0)= 11
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7
f (0,5)= 8 f (3,0)= 11
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)=
8 f (3,0)= 11
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8
f (3,0)= 11
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)=
11
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary x = 0:
g (y)= f (0, y) = 3y +7g (y) = 3
no critical pointsAt boundary y = 0:
g (x)= f (x,0) = 6x+7g (x) = 6
no critical points
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary x = 0:
g (y)=
f (0, y) = 3y +7g (y) = 3
no critical pointsAt boundary y = 0:
g (x)= f (x,0) = 6x+7g (x) = 6
no critical points
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary x = 0:
g (y)= f (0, y) =
3y +7g (y) = 3
no critical pointsAt boundary y = 0:
g (x)= f (x,0) = 6x+7g (x) = 6
no critical points
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary x = 0:
g (y)= f (0, y) = 3y +7
g (y) = 3
no critical pointsAt boundary y = 0:
g (x)= f (x,0) = 6x+7g (x) = 6
no critical points
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary x = 0:
g (y)= f (0, y) = 3y +7g (y) = 3
no critical pointsAt boundary y = 0:
g (x)= f (x,0) = 6x+7g (x) = 6
no critical points
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary x = 0:
g (y)= f (0, y) = 3y +7g (y) = 3
no critical points
At boundary y = 0:
g (x)= f (x,0) = 6x+7g (x) = 6
no critical points
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary x = 0:
g (y)= f (0, y) = 3y +7g (y) = 3
no critical pointsAt boundary y = 0:
g (x)= f (x,0) = 6x+7g (x) = 6
no critical points
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary x = 0:
g (y)= f (0, y) = 3y +7g (y) = 3
no critical pointsAt boundary y = 0:
g (x)=
f (x,0) = 6x+7g (x) = 6
no critical points
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary x = 0:
g (y)= f (0, y) = 3y +7g (y) = 3
no critical pointsAt boundary y = 0:
g (x)= f (x,0) =
6x+7g (x) = 6
no critical points
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary x = 0:
g (y)= f (0, y) = 3y +7g (y) = 3
no critical pointsAt boundary y = 0:
g (x)= f (x,0) = 6x+7
g (x) = 6
no critical points
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary x = 0:
g (y)= f (0, y) = 3y +7g (y) = 3
no critical pointsAt boundary y = 0:
g (x)= f (x,0) = 6x+7g (x) = 6
no critical points
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary x = 0:
g (y)= f (0, y) = 3y +7g (y) = 3
no critical pointsAt boundary y = 0:
g (x)= f (x,0) = 6x+7g (x) = 6
no critical points
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary y =53x+5:
g (x) = f(x,5
3x+5
)= 3x
(53x+5
)6x3
(53x+5
)+7
= 5x2+24x8g (x) = 10x+24
Critical points:
10x+24= 0 x = 125
& y = 1
f(125 ,1
)= 20.8
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary y =53x+5:
g (x) = f(x,5
3x+5
)= 3x
(53x+5
)6x3
(53x+5
)+7
= 5x2+24x8g (x) = 10x+24
Critical points:
10x+24= 0 x = 125
& y = 1
f(125 ,1
)= 20.8
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary y =53x+5:
g (x) = f(x,5
3x+5
)
= 3x(53x+5
)6x3
(53x+5
)+7
= 5x2+24x8g (x) = 10x+24
Critical points:
10x+24= 0 x = 125
& y = 1
f(125 ,1
)= 20.8
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary y =53x+5:
g (x) = f(x,5
3x+5
)= 3x
(53x+5
)6x3
(53x+5
)+7
= 5x2+24x8g (x) = 10x+24
Critical points:
10x+24= 0 x = 125
& y = 1
f(125 ,1
)= 20.8
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary y =53x+5:
g (x) = f(x,5
3x+5
)= 3x
(53x+5
)6x3
(53x+5
)+7
= 5x2+24x8
g (x) = 10x+24
Critical points:
10x+24= 0 x = 125
& y = 1
f(125 ,1
)= 20.8
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary y =53x+5:
g (x) = f(x,5
3x+5
)= 3x
(53x+5
)6x3
(53x+5
)+7
= 5x2+24x8g (x) = 10x+24
Critical points:
10x+24= 0 x = 125
& y = 1
f(125 ,1
)= 20.8
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary y =53x+5:
g (x) = f(x,5
3x+5
)= 3x
(53x+5
)6x3
(53x+5
)+7
= 5x2+24x8g (x) = 10x+24
Critical points:
10x+24= 0 x = 125
& y = 1
f(125 ,1
)= 20.8
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary y =53x+5:
g (x) = f(x,5
3x+5
)= 3x
(53x+5
)6x3
(53x+5
)+7
= 5x2+24x8g (x) = 10x+24
Critical points:
10x+24= 0
x = 125
& y = 1
f(125 ,1
)= 20.8
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary y =53x+5:
g (x) = f(x,5
3x+5
)= 3x
(53x+5
)6x3
(53x+5
)+7
= 5x2+24x8g (x) = 10x+24
Critical points:
10x+24= 0 x = 125
&
y = 1
f(125 ,1
)= 20.8
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary y =53x+5:
g (x) = f(x,5
3x+5
)= 3x
(53x+5
)6x3
(53x+5
)+7
= 5x2+24x8g (x) = 10x+24
Critical points:
10x+24= 0 x = 125
& y = 1
f(125 ,1
)= 20.8
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary y =53x+5:
g (x) = f(x,5
3x+5
)= 3x
(53x+5
)6x3
(53x+5
)+7
= 5x2+24x8g (x) = 10x+24
Critical points:
10x+24= 0 x = 125
& y = 1
f(125 ,1
)= 20.8
ExampleGiven: f (x, y)= 3xy 6x3y +7
At critical point: f (1,2)= 1At vertices: f (0,0)= 7 f (0,5)=8 f (3,0)=11At boundary y =53x+5: f
(125 ,1
)= 20.8
CONCLUSION: The function has an absolute maximum at(125 ,1
)and an
absolute minimum at (3,0) with values 20.8 and 11, respectively.
ExampleGiven: f (x, y)= 3xy 6x3y +7
At critical point: f (1,2)= 1At vertices: f (0,0)= 7 f (0,5)=8 f (3,0)=11At boundary y =53x+5: f
(125 ,1
)= 20.8CONCLUSION: The function has an absolute maximum at
(125 ,1
)and an
absolute minimum at (3,0) with values 20.8 and 11, respectively.
RemarkLet f be a function of two variables and R = {(x, y) | x > 0 and y > 0}.Suppose f is continuous on R and (x0, y0) R is the only critical point off .
1 If f (x, y) is very large when either x or y are very close to zero or verylarge, then f has an absolute minimum at (x0, y0).
2 If f (x, y) is "very negative" when either x or y are very close to zeroor very large, then f has an absolute maximum at (x0, y0).
RemarkLet f be a function of two variables and R = {(x, y) | x > 0 and y > 0}.Suppose f is continuous on R and (x0, y0) R is the only critical point off .
1 If f (x, y) is very large when either x or y are very close to zero or verylarge, then f has an absolute minimum at (x0, y0).
2 If f (x, y) is "very negative" when either x or y are very close to zeroor very large, then f has an absolute maximum at (x0, y0).
RemarkLet f be a function of two variables and R = {(x, y) | x > 0 and y > 0}.Suppose f is continuous on R and (x0, y0) R is the only critical point off .
1 If f (x, y) is very large when either x or y are very close to zero or verylarge, then f has an absolute minimum at (x0, y0).
2 If f (x, y) is "very negative" when either x or y are very close to zeroor very large, then f has an absolute maximum at (x0, y0).
RemarkLet f be a function of two variables and R = {(x, y) | x > 0 and y > 0}.Suppose f is continuous on R and (x0, y0) R is the only critical point off .
1 If f (x, y) is very large when either x or y are very close to zero or verylarge, then f has an absolute minimum at (x0, y0).
2 If f (x, y) is "very negative" when either x or y are very close to zeroor very large, then f has an absolute maximum at (x0, y0).
ExampleA manufacturer wishes to make an open rectangular box with volume500 cm3 using the least possible amount of material. Find the dimensionsof the box.
Minimize: Surface AreaConstraint: V = 500 cm3ORMinimize: A = lw +2wh+2lhConstraint: lwh = 500ORMinimize: A(l ,w)= lw + 1000
l+ 1000
wRegion: {(l ,w) | l > 0, w > 0}
ExampleA manufacturer wishes to make an open rectangular box with volume500 cm3 using the least possible amount of material. Find the dimensionsof the box.
Minimize:
Surface AreaConstraint: V = 500 cm3ORMinimize: A = lw +2wh+2lhConstraint: lwh = 500ORMinimize: A(l ,w)= lw + 1000
l+ 1000
wRegion: {(l ,w) | l > 0, w > 0}
ExampleA manufacturer wishes to make an open rectangular box with volume500 cm3 using the least possible amount of material. Find the dimensionsof the box.
Minimize: Surface Area
Constraint: V = 500 cm3ORMinimize: A = lw +2wh+2lhConstraint: lwh = 500ORMinimize: A(l ,w)= lw + 1000
l+ 1000
wRegion: {(l ,w) | l > 0, w > 0}
ExampleA manufacturer wishes to make an open rectangular box with volume500 cm3 using the least possible amount of material. Find the dimensionsof the box.
Minimize: Surface AreaConstraint:
V = 500 cm3ORMinimize: A = lw +2wh+2lhConstraint: lwh = 500ORMinimize: A(l ,w)= lw + 1000
l+ 1000
wRegion: {(l ,w) | l > 0, w > 0}
ExampleA manufacturer wishes to make an open rectangular box with volume500 cm3 using the least possible amount of material. Find the dimensionsof the box.
Minimize: Surface AreaConstraint: V = 500 cm3
ORMinimize: A = lw +2wh+2lhConstraint: lwh = 500ORMinimize: A(l ,w)= lw + 1000
l+ 1000
wRegion: {(l ,w) | l > 0, w > 0}
ExampleA manufacturer wishes to make an open rectangular box with volume500 cm3 using the least possible amount of material. Find the dimensionsof the box.
Minimize: Surface AreaConstraint: V = 500 cm3ORMinimize: A = lw +2wh+2lhConstraint: lwh = 500
ORMinimize: A(l ,w)= lw + 1000
l+ 1000
wRegion: {(l ,w) | l > 0, w > 0}
ExampleA manufacturer wishes to make an open rectangular box with volume500 cm3 using the least possible amount of material. Find the dimensionsof the box.
Minimize: Surface AreaConstraint: V = 500 cm3ORMinimize: A = lw +2wh+2lhConstraint: lwh = 500ORMinimize: A(l ,w)= lw + 1000
l+ 1000
wRegion: {(l ,w) | l > 0, w > 0}
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Al (l ,w)=w 1000
l2and Aw (l ,w)= l 1000
w2
Critical Point: l = 10 and w = 10.
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Al (l ,w)=
w 1000l2
and Aw (l ,w)= l 1000w2
Critical Point: l = 10 and w = 10.
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Al (l ,w)=w
1000l2
and Aw (l ,w)= l 1000w2
Critical Point: l = 10 and w = 10.
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Al (l ,w)=w 1000
l2
and Aw (l ,w)= l 1000w2
Critical Point: l = 10 and w = 10.
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Al (l ,w)=w 1000
l2and
Aw (l ,w)= l 1000w2
Critical Point: l = 10 and w = 10.
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Al (l ,w)=w 1000
l2and Aw (l ,w)=
l 1000w2
Critical Point: l = 10 and w = 10.
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Al (l ,w)=w 1000
l2and Aw (l ,w)= l
1000w2
Critical Point: l = 10 and w = 10.
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Al (l ,w)=w 1000
l2and Aw (l ,w)= l 1000
w2
Critical Point: l = 10 and w = 10.
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Al (l ,w)=w 1000
l2and Aw (l ,w)= l 1000
w2
Critical Point: l = 10 and w = 10.
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.
Hold w constant:liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:
limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.
The minimum surface area is 300cm2.
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
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