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Section 4.5Optimization Problems
V63.0121.002.2010Su, Calculus I
New York University
June 14, 2010
Announcements
I The midterm is graded!I Quiz 4 Thursday on 4.1–4.4I Guest speaker on Thursday: Arjun Krishnan
. . . . . .
. . . . . .
Announcements
I The midterm is graded!I Quiz 4 Thursday on
4.1–4.4I Guest speaker on
Thursday: Arjun Krishnan
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 2 / 31
. . . . . .
Objectives
I Given a problem requiringoptimization, identify theobjective functions,variables, and constraints.
I Solve optimizationproblems with calculus.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 3 / 31
. . . . . .
Outline
Leading by Example
The Text in the Box
More Examples
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 4 / 31
. . . . . .
Leading by Example
Example
What is the rectangle of fixed perimeter with maximum area?
Solution
I Draw a rectangle.
.
.ℓ
.w
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 5 / 31
. . . . . .
Leading by Example
Example
What is the rectangle of fixed perimeter with maximum area?
Solution
I Draw a rectangle.
.
.ℓ
.w
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 5 / 31
. . . . . .
Leading by Example
Example
What is the rectangle of fixed perimeter with maximum area?
Solution
I Draw a rectangle.
..ℓ
.w
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 5 / 31
. . . . . .
Leading by Example
Example
What is the rectangle of fixed perimeter with maximum area?
Solution
I Draw a rectangle.
..ℓ
.w
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 5 / 31
. . . . . .
Solution Continued
I Let its length be ℓ and its width be w. The objective function isarea A = ℓw.
I This is a function of two variables, not one. But the perimeter isfixed.
I Since p = 2ℓ+ 2w, we have ℓ =p− 2w
2, so
A = ℓw =p− 2w
2· w =
12(p− 2w)(w) =
12pw− w2
I Now we have A as a function of w alone (p is constant).I The natural domain of this function is [0,p/2] (we want to make
sure A(w) ≥ 0).
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 6 / 31
. . . . . .
Solution Continued
I Let its length be ℓ and its width be w. The objective function isarea A = ℓw.
I This is a function of two variables, not one. But the perimeter isfixed.
I Since p = 2ℓ+ 2w, we have ℓ =p− 2w
2, so
A = ℓw =p− 2w
2· w =
12(p− 2w)(w) =
12pw− w2
I Now we have A as a function of w alone (p is constant).I The natural domain of this function is [0,p/2] (we want to make
sure A(w) ≥ 0).
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 6 / 31
. . . . . .
Solution Continued
I Let its length be ℓ and its width be w. The objective function isarea A = ℓw.
I This is a function of two variables, not one. But the perimeter isfixed.
I Since p = 2ℓ+ 2w, we have ℓ =p− 2w
2,
so
A = ℓw =p− 2w
2· w =
12(p− 2w)(w) =
12pw− w2
I Now we have A as a function of w alone (p is constant).I The natural domain of this function is [0,p/2] (we want to make
sure A(w) ≥ 0).
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 6 / 31
. . . . . .
Solution Continued
I Let its length be ℓ and its width be w. The objective function isarea A = ℓw.
I This is a function of two variables, not one. But the perimeter isfixed.
I Since p = 2ℓ+ 2w, we have ℓ =p− 2w
2, so
A = ℓw =p− 2w
2· w =
12(p− 2w)(w) =
12pw− w2
I Now we have A as a function of w alone (p is constant).I The natural domain of this function is [0,p/2] (we want to make
sure A(w) ≥ 0).
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 6 / 31
. . . . . .
Solution Continued
I Let its length be ℓ and its width be w. The objective function isarea A = ℓw.
I This is a function of two variables, not one. But the perimeter isfixed.
I Since p = 2ℓ+ 2w, we have ℓ =p− 2w
2, so
A = ℓw =p− 2w
2· w =
12(p− 2w)(w) =
12pw− w2
I Now we have A as a function of w alone (p is constant).
I The natural domain of this function is [0,p/2] (we want to makesure A(w) ≥ 0).
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 6 / 31
. . . . . .
Solution Continued
I Let its length be ℓ and its width be w. The objective function isarea A = ℓw.
I This is a function of two variables, not one. But the perimeter isfixed.
I Since p = 2ℓ+ 2w, we have ℓ =p− 2w
2, so
A = ℓw =p− 2w
2· w =
12(p− 2w)(w) =
12pw− w2
I Now we have A as a function of w alone (p is constant).I The natural domain of this function is [0,p/2] (we want to make
sure A(w) ≥ 0).
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 6 / 31
. . . . . .
Solution Concluded
We use the Closed Interval Method for A(w) =12pw− w2 on [0,p/2].
I At the endpoints, A(0) = A(p/2) = 0.
I To find the critical points, we finddAdw
=12p− 2w.
I The critical points are when
0 =12p− 2w =⇒ w =
p4
I Since this is the only critical point, it must be the maximum. In thiscase ℓ =
p4as well.
I We have a square! The maximal area is A(p/4) = p2/16.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 7 / 31
. . . . . .
Solution Concluded
We use the Closed Interval Method for A(w) =12pw− w2 on [0,p/2].
I At the endpoints, A(0) = A(p/2) = 0.
I To find the critical points, we finddAdw
=12p− 2w.
I The critical points are when
0 =12p− 2w =⇒ w =
p4
I Since this is the only critical point, it must be the maximum. In thiscase ℓ =
p4as well.
I We have a square! The maximal area is A(p/4) = p2/16.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 7 / 31
. . . . . .
Solution Concluded
We use the Closed Interval Method for A(w) =12pw− w2 on [0,p/2].
I At the endpoints, A(0) = A(p/2) = 0.
I To find the critical points, we finddAdw
=12p− 2w.
I The critical points are when
0 =12p− 2w =⇒ w =
p4
I Since this is the only critical point, it must be the maximum. In thiscase ℓ =
p4as well.
I We have a square! The maximal area is A(p/4) = p2/16.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 7 / 31
. . . . . .
Solution Concluded
We use the Closed Interval Method for A(w) =12pw− w2 on [0,p/2].
I At the endpoints, A(0) = A(p/2) = 0.
I To find the critical points, we finddAdw
=12p− 2w.
I The critical points are when
0 =12p− 2w =⇒ w =
p4
I Since this is the only critical point, it must be the maximum. In thiscase ℓ =
p4as well.
I We have a square! The maximal area is A(p/4) = p2/16.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 7 / 31
. . . . . .
Solution Concluded
We use the Closed Interval Method for A(w) =12pw− w2 on [0,p/2].
I At the endpoints, A(0) = A(p/2) = 0.
I To find the critical points, we finddAdw
=12p− 2w.
I The critical points are when
0 =12p− 2w =⇒ w =
p4
I Since this is the only critical point, it must be the maximum. In thiscase ℓ =
p4as well.
I We have a square! The maximal area is A(p/4) = p2/16.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 7 / 31
. . . . . .
Outline
Leading by Example
The Text in the Box
More Examples
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 8 / 31
. . . . . .
Strategies for Problem Solving
1. Understand the problem2. Devise a plan3. Carry out the plan4. Review and extend
György Pólya(Hungarian, 1887–1985)
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 9 / 31
. . . . . .
The Text in the Box
1. Understand the Problem. What is known? What is unknown?What are the conditions?
2. Draw a diagram.3. Introduce Notation.4. Express the “objective function” Q in terms of the other symbols5. If Q is a function of more than one “decision variable”, use the
given information to eliminate all but one of them.6. Find the absolute maximum (or minimum, depending on the
problem) of the function on its domain.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 10 / 31
. . . . . .
The Text in the Box
1. Understand the Problem. What is known? What is unknown?What are the conditions?
2. Draw a diagram.
3. Introduce Notation.4. Express the “objective function” Q in terms of the other symbols5. If Q is a function of more than one “decision variable”, use the
given information to eliminate all but one of them.6. Find the absolute maximum (or minimum, depending on the
problem) of the function on its domain.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 10 / 31
. . . . . .
The Text in the Box
1. Understand the Problem. What is known? What is unknown?What are the conditions?
2. Draw a diagram.3. Introduce Notation.
4. Express the “objective function” Q in terms of the other symbols5. If Q is a function of more than one “decision variable”, use the
given information to eliminate all but one of them.6. Find the absolute maximum (or minimum, depending on the
problem) of the function on its domain.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 10 / 31
. . . . . .
The Text in the Box
1. Understand the Problem. What is known? What is unknown?What are the conditions?
2. Draw a diagram.3. Introduce Notation.4. Express the “objective function” Q in terms of the other symbols
5. If Q is a function of more than one “decision variable”, use thegiven information to eliminate all but one of them.
6. Find the absolute maximum (or minimum, depending on theproblem) of the function on its domain.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 10 / 31
. . . . . .
The Text in the Box
1. Understand the Problem. What is known? What is unknown?What are the conditions?
2. Draw a diagram.3. Introduce Notation.4. Express the “objective function” Q in terms of the other symbols5. If Q is a function of more than one “decision variable”, use the
given information to eliminate all but one of them.
6. Find the absolute maximum (or minimum, depending on theproblem) of the function on its domain.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 10 / 31
. . . . . .
The Text in the Box
1. Understand the Problem. What is known? What is unknown?What are the conditions?
2. Draw a diagram.3. Introduce Notation.4. Express the “objective function” Q in terms of the other symbols5. If Q is a function of more than one “decision variable”, use the
given information to eliminate all but one of them.6. Find the absolute maximum (or minimum, depending on the
problem) of the function on its domain.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 10 / 31
. . . . . .
Polya's Method in Kindergarten
Name [_
Problem Solving StrategyDraw a Picture
Kathy had a box of 8 crayons.She gave some crayons away.She has 5 left.How many crayons did Kathy give away?
UNDERSTAND•
What do you want to find out?Draw a line under the question.
You can draw a pictureto solve the problem.
crayons
What number do Iadd to 5 to get 8?
8 - = 55 + 3 = 8
CHECK
Does your answer make sense?Explain.
Draw a picture to solve the problem.Write how many were given away.I. I had 10 pencils.
I gave some away.I have 3 left. How manypencils did I give away?
~7
What numberdo I add to 3to make 10?
13iftill:ii ?11
ftI'•'«II
ft A
H 11M i l
U U U U> U U
I I
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 11 / 31
. . . . . .
Recall: The Closed Interval MethodSee Section 4.1
The Closed Interval MethodTo find the extreme values of a function f on [a,b], we need to:
I Evaluate f at the endpoints a and bI Evaluate f at the critical points x where either f′(x) = 0 or f is not
differentiable at x.I The points with the largest function value are the global maximum
pointsI The points with the smallest/most negative function value are the
global minimum points.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 12 / 31
. . . . . .
Recall: The First Derivative TestSee Section 4.3
Theorem (The First Derivative Test)
Let f be continuous on (a,b) and c a critical point of f in (a,b).I If f′ changes from negative to positive at c, then c is a local
minimum.I If f′ changes from positive to negative at c, then c is a local
maximum.I If f′ does not change sign at c, then c is not a local extremum.
Corollary
I If f′ < 0 for all x < c and f′(x) > 0 for all x > c, then c is the globalminimum of f on (a,b).
I If f′ < 0 for all x > c and f′(x) > 0 for all x < c, then c is the globalmaximum of f on (a,b).
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 13 / 31
. . . . . .
Recall: The First Derivative TestSee Section 4.3
Theorem (The First Derivative Test)
Let f be continuous on (a,b) and c a critical point of f in (a,b).I If f′ changes from negative to positive at c, then c is a local
minimum.I If f′ changes from positive to negative at c, then c is a local
maximum.I If f′ does not change sign at c, then c is not a local extremum.
Corollary
I If f′ < 0 for all x < c and f′(x) > 0 for all x > c, then c is the globalminimum of f on (a,b).
I If f′ < 0 for all x > c and f′(x) > 0 for all x < c, then c is the globalmaximum of f on (a,b).
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 13 / 31
. . . . . .
Recall: The Second Derivative TestSee Section 4.3
Theorem (The Second Derivative Test)
Let f, f′, and f′′ be continuous on [a,b]. Let c be in (a,b) with f′(c) = 0.I If f′′(c) < 0, then f(c) is a local maximum.I If f′′(c) > 0, then f(c) is a local minimum.
Warning
If f′′(c) = 0, the second derivative test is inconclusive (this does notmean c is neither; we just don’t know yet).
Corollary
I If f′(c) = 0 and f′′(x) > 0 for all x, then c is the global minimum of fI If f′(c) = 0 and f′′(x) < 0 for all x, then c is the global maximum of f
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 14 / 31
. . . . . .
Recall: The Second Derivative TestSee Section 4.3
Theorem (The Second Derivative Test)
Let f, f′, and f′′ be continuous on [a,b]. Let c be in (a,b) with f′(c) = 0.I If f′′(c) < 0, then f(c) is a local maximum.I If f′′(c) > 0, then f(c) is a local minimum.
Warning
If f′′(c) = 0, the second derivative test is inconclusive (this does notmean c is neither; we just don’t know yet).
Corollary
I If f′(c) = 0 and f′′(x) > 0 for all x, then c is the global minimum of fI If f′(c) = 0 and f′′(x) < 0 for all x, then c is the global maximum of f
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 14 / 31
. . . . . .
Recall: The Second Derivative TestSee Section 4.3
Theorem (The Second Derivative Test)
Let f, f′, and f′′ be continuous on [a,b]. Let c be in (a,b) with f′(c) = 0.I If f′′(c) < 0, then f(c) is a local maximum.I If f′′(c) > 0, then f(c) is a local minimum.
Warning
If f′′(c) = 0, the second derivative test is inconclusive (this does notmean c is neither; we just don’t know yet).
Corollary
I If f′(c) = 0 and f′′(x) > 0 for all x, then c is the global minimum of fI If f′(c) = 0 and f′′(x) < 0 for all x, then c is the global maximum of f
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 14 / 31
. . . . . .
Which to use when?
CIM 1DT 2DTPro – no need for
inequalities– gets globalextremaautomatically
– works onnon-closed,non-boundedintervals– only one derivative
– works onnon-closed,non-boundedintervals– no need forinequalities
Con – only for closedbounded intervals
– Uses inequalities– More work atboundary than CIM
– More derivatives– less conclusivethan 1DT– more work atboundary than CIM
I Use CIM if it applies: the domain is a closed, bounded intervalI If domain is not closed or not bounded, use 2DT if you like to take
derivatives, or 1DT if you like to compare signs.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 15 / 31
. . . . . .
Which to use when?
CIM 1DT 2DTPro – no need for
inequalities– gets globalextremaautomatically
– works onnon-closed,non-boundedintervals– only one derivative
– works onnon-closed,non-boundedintervals– no need forinequalities
Con – only for closedbounded intervals
– Uses inequalities– More work atboundary than CIM
– More derivatives– less conclusivethan 1DT– more work atboundary than CIM
I Use CIM if it applies: the domain is a closed, bounded intervalI If domain is not closed or not bounded, use 2DT if you like to take
derivatives, or 1DT if you like to compare signs.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 15 / 31
. . . . . .
Outline
Leading by Example
The Text in the Box
More Examples
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 16 / 31
. . . . . .
Another Example
Example (The Best Fencing Plan)
A rectangular plot of farmland will be bounded on one side by a riverand on the other three sides by a single-strand electric fence. With800m of wire at your disposal, what is the largest area you canenclose, and what are its dimensions?
I Known: amount of fence usedI Unknown: area enclosedI Objective: maximize areaI Constraint: fixed fence length
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 17 / 31
. . . . . .
Solution
1. Everybody understand?
2. Draw a diagram.3. Length and width are ℓ and w. Length of wire used is p.4. Q = area = ℓw.5. Since p = ℓ+ 2w, we have ℓ = p− 2w and so
Q(w) = (p− 2w)(w) = pw− 2w2
The domain of Q is [0,p/2]
6. dQdw
= p− 4w, which is zero when w =p4.
Q(0) = Q(p/2) = 0, but
Q(p4
)= p · p
4− 2 · p
2
16=
p2
8= 80,000m2
so the critical point is the absolute maximum.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 18 / 31
. . . . . .
Another Example
Example (The Best Fencing Plan)
A rectangular plot of farmland will be bounded on one side by a riverand on the other three sides by a single-strand electric fence. With800m of wire at your disposal, what is the largest area you canenclose, and what are its dimensions?
I Known: amount of fence usedI Unknown: area enclosedI Objective: maximize areaI Constraint: fixed fence length
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 19 / 31
. . . . . .
Another Example
Example (The Best Fencing Plan)
A rectangular plot of farmland will be bounded on one side by a riverand on the other three sides by a single-strand electric fence. With800m of wire at your disposal, what is the largest area you canenclose, and what are its dimensions?
I Known: amount of fence usedI Unknown: area enclosed
I Objective: maximize areaI Constraint: fixed fence length
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 19 / 31
. . . . . .
Another Example
Example (The Best Fencing Plan)
A rectangular plot of farmland will be bounded on one side by a riverand on the other three sides by a single-strand electric fence. With800m of wire at your disposal, what is the largest area you canenclose, and what are its dimensions?
I Known: amount of fence usedI Unknown: area enclosedI Objective: maximize areaI Constraint: fixed fence length
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 19 / 31
. . . . . .
Solution
1. Everybody understand?
2. Draw a diagram.3. Length and width are ℓ and w. Length of wire used is p.4. Q = area = ℓw.5. Since p = ℓ+ 2w, we have ℓ = p− 2w and so
Q(w) = (p− 2w)(w) = pw− 2w2
The domain of Q is [0,p/2]
6. dQdw
= p− 4w, which is zero when w =p4.
Q(0) = Q(p/2) = 0, but
Q(p4
)= p · p
4− 2 · p
2
16=
p2
8= 80,000m2
so the critical point is the absolute maximum.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 20 / 31
. . . . . .
Solution
1. Everybody understand?2. Draw a diagram.
3. Length and width are ℓ and w. Length of wire used is p.4. Q = area = ℓw.5. Since p = ℓ+ 2w, we have ℓ = p− 2w and so
Q(w) = (p− 2w)(w) = pw− 2w2
The domain of Q is [0,p/2]
6. dQdw
= p− 4w, which is zero when w =p4.
Q(0) = Q(p/2) = 0, but
Q(p4
)= p · p
4− 2 · p
2
16=
p2
8= 80,000m2
so the critical point is the absolute maximum.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 20 / 31
. . . . . .
Diagram
A rectangular plot of farmland will be bounded on one side by a riverand on the other three sides by a single-strand electric fence. With 800m of wire at your disposal, what is the largest area you can enclose,and what are its dimensions?
.
.
.
.
.w
.ℓ
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 21 / 31
. . . . . .
Solution
1. Everybody understand?2. Draw a diagram.
3. Length and width are ℓ and w. Length of wire used is p.4. Q = area = ℓw.5. Since p = ℓ+ 2w, we have ℓ = p− 2w and so
Q(w) = (p− 2w)(w) = pw− 2w2
The domain of Q is [0,p/2]
6. dQdw
= p− 4w, which is zero when w =p4.
Q(0) = Q(p/2) = 0, but
Q(p4
)= p · p
4− 2 · p
2
16=
p2
8= 80,000m2
so the critical point is the absolute maximum.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 22 / 31
. . . . . .
Solution
1. Everybody understand?2. Draw a diagram.3. Length and width are ℓ and w. Length of wire used is p.
4. Q = area = ℓw.5. Since p = ℓ+ 2w, we have ℓ = p− 2w and so
Q(w) = (p− 2w)(w) = pw− 2w2
The domain of Q is [0,p/2]
6. dQdw
= p− 4w, which is zero when w =p4.
Q(0) = Q(p/2) = 0, but
Q(p4
)= p · p
4− 2 · p
2
16=
p2
8= 80,000m2
so the critical point is the absolute maximum.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 22 / 31
. . . . . .
Diagram
A rectangular plot of farmland will be bounded on one side by a riverand on the other three sides by a single-strand electric fence. With 800m of wire at your disposal, what is the largest area you can enclose,and what are its dimensions?
.
.
.
.
.w
.ℓ
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 23 / 31
. . . . . .
Diagram
A rectangular plot of farmland will be bounded on one side by a riverand on the other three sides by a single-strand electric fence. With 800m of wire at your disposal, what is the largest area you can enclose,and what are its dimensions?
.
.
.
.
.w
.ℓ
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 23 / 31
. . . . . .
Solution
1. Everybody understand?2. Draw a diagram.3. Length and width are ℓ and w. Length of wire used is p.
4. Q = area = ℓw.5. Since p = ℓ+ 2w, we have ℓ = p− 2w and so
Q(w) = (p− 2w)(w) = pw− 2w2
The domain of Q is [0,p/2]
6. dQdw
= p− 4w, which is zero when w =p4.
Q(0) = Q(p/2) = 0, but
Q(p4
)= p · p
4− 2 · p
2
16=
p2
8= 80,000m2
so the critical point is the absolute maximum.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 24 / 31
. . . . . .
Solution
1. Everybody understand?2. Draw a diagram.3. Length and width are ℓ and w. Length of wire used is p.4. Q = area = ℓw.
5. Since p = ℓ+ 2w, we have ℓ = p− 2w and so
Q(w) = (p− 2w)(w) = pw− 2w2
The domain of Q is [0,p/2]
6. dQdw
= p− 4w, which is zero when w =p4.
Q(0) = Q(p/2) = 0, but
Q(p4
)= p · p
4− 2 · p
2
16=
p2
8= 80,000m2
so the critical point is the absolute maximum.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 24 / 31
. . . . . .
Solution
1. Everybody understand?2. Draw a diagram.3. Length and width are ℓ and w. Length of wire used is p.4. Q = area = ℓw.5. Since p = ℓ+ 2w, we have ℓ = p− 2w and so
Q(w) = (p− 2w)(w) = pw− 2w2
The domain of Q is [0,p/2]
6. dQdw
= p− 4w, which is zero when w =p4.
Q(0) = Q(p/2) = 0, but
Q(p4
)= p · p
4− 2 · p
2
16=
p2
8= 80,000m2
so the critical point is the absolute maximum.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 24 / 31
. . . . . .
Solution
1. Everybody understand?2. Draw a diagram.3. Length and width are ℓ and w. Length of wire used is p.4. Q = area = ℓw.5. Since p = ℓ+ 2w, we have ℓ = p− 2w and so
Q(w) = (p− 2w)(w) = pw− 2w2
The domain of Q is [0,p/2]
6. dQdw
= p− 4w, which is zero when w =p4.
Q(0) = Q(p/2) = 0, but
Q(p4
)= p · p
4− 2 · p
2
16=
p2
8= 80,000m2
so the critical point is the absolute maximum.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 24 / 31
. . . . . .
Solution
1. Everybody understand?2. Draw a diagram.3. Length and width are ℓ and w. Length of wire used is p.4. Q = area = ℓw.5. Since p = ℓ+ 2w, we have ℓ = p− 2w and so
Q(w) = (p− 2w)(w) = pw− 2w2
The domain of Q is [0,p/2]
6. dQdw
= p− 4w, which is zero when w =p4.
Q(0) = Q(p/2) = 0, but
Q(p4
)= p · p
4− 2 · p
2
16=
p2
8= 80,000m2
so the critical point is the absolute maximum.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 24 / 31
. . . . . .
Solution
1. Everybody understand?2. Draw a diagram.3. Length and width are ℓ and w. Length of wire used is p.4. Q = area = ℓw.5. Since p = ℓ+ 2w, we have ℓ = p− 2w and so
Q(w) = (p− 2w)(w) = pw− 2w2
The domain of Q is [0,p/2]
6. dQdw
= p− 4w, which is zero when w =p4. Q(0) = Q(p/2) = 0, but
Q(p4
)= p · p
4− 2 · p
2
16=
p2
8= 80,000m2
so the critical point is the absolute maximum.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 24 / 31
. . . . . .
Your turn
Example (The shortest fence)
A 216m2 rectangular pea patch is to be enclosed by a fence anddivided into two equal parts by another fence parallel to one of itssides. What dimensions for the outer rectangle will require the smallesttotal length of fence? How much fence will be needed?
SolutionLet the length and width of the pea patch be ℓ and w. The amount offence needed is f = 2ℓ+ 3w. Since ℓw = A, a constant, we have
f(w) = 2Aw
+ 3w.
The domain is all positive numbers.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 25 / 31
. . . . . .
Your turn
Example (The shortest fence)
A 216m2 rectangular pea patch is to be enclosed by a fence anddivided into two equal parts by another fence parallel to one of itssides. What dimensions for the outer rectangle will require the smallesttotal length of fence? How much fence will be needed?
SolutionLet the length and width of the pea patch be ℓ and w. The amount offence needed is f = 2ℓ+ 3w. Since ℓw = A, a constant, we have
f(w) = 2Aw
+ 3w.
The domain is all positive numbers.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 25 / 31
. . . . . .
Diagram
.
. .
.ℓ
.w
f = 2ℓ+ 3w A = ℓw ≡ 216
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 26 / 31
. . . . . .
Solution (Continued)
We need to find the minimum value of f(w) =2Aw
+ 3w on (0,∞).
I We havedfdw
= −2Aw2 + 3
which is zero when w =
√2A3.
I Since f′′(w) = 4Aw−3, which is positive for all positive w, thecritical point is a minimum, in fact the global minimum.
I So the area is minimized when w =
√2A3
= 12 and
ℓ =Aw
=
√3A2
= 18. The amount of fence needed is
f
(√2A3
)= 2 ·
√3A2
+ 3√
2A3
= 2√6A = 2
√6 · 216 = 72m
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 27 / 31
. . . . . .
Solution (Continued)
We need to find the minimum value of f(w) =2Aw
+ 3w on (0,∞).I We have
dfdw
= −2Aw2 + 3
which is zero when w =
√2A3.
I Since f′′(w) = 4Aw−3, which is positive for all positive w, thecritical point is a minimum, in fact the global minimum.
I So the area is minimized when w =
√2A3
= 12 and
ℓ =Aw
=
√3A2
= 18. The amount of fence needed is
f
(√2A3
)= 2 ·
√3A2
+ 3√
2A3
= 2√6A = 2
√6 · 216 = 72m
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 27 / 31
. . . . . .
Solution (Continued)
We need to find the minimum value of f(w) =2Aw
+ 3w on (0,∞).I We have
dfdw
= −2Aw2 + 3
which is zero when w =
√2A3.
I Since f′′(w) = 4Aw−3, which is positive for all positive w, thecritical point is a minimum, in fact the global minimum.
I So the area is minimized when w =
√2A3
= 12 and
ℓ =Aw
=
√3A2
= 18. The amount of fence needed is
f
(√2A3
)= 2 ·
√3A2
+ 3√
2A3
= 2√6A = 2
√6 · 216 = 72m
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 27 / 31
. . . . . .
Solution (Continued)
We need to find the minimum value of f(w) =2Aw
+ 3w on (0,∞).I We have
dfdw
= −2Aw2 + 3
which is zero when w =
√2A3.
I Since f′′(w) = 4Aw−3, which is positive for all positive w, thecritical point is a minimum, in fact the global minimum.
I So the area is minimized when w =
√2A3
= 12 and
ℓ =Aw
=
√3A2
= 18. The amount of fence needed is
f
(√2A3
)= 2 ·
√3A2
+ 3√
2A3
= 2√6A = 2
√6 · 216 = 72m
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 27 / 31
. . . . . .
Try this one
Example
An advertisement consists of a rectangular printed region plus 1 inmargins on the sides and 1.5 in margins on the top and bottom. If thetotal area of the advertisement is to be 120 in2, what dimensions shouldthe advertisement be to maximize the area of the printed region?
AnswerThe optimal paper dimensions are 4
√5 in by 6
√5 in.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 28 / 31
. . . . . .
Try this one
Example
An advertisement consists of a rectangular printed region plus 1 inmargins on the sides and 1.5 in margins on the top and bottom. If thetotal area of the advertisement is to be 120 in2, what dimensions shouldthe advertisement be to maximize the area of the printed region?
AnswerThe optimal paper dimensions are 4
√5 in by 6
√5 in.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 28 / 31
. . . . . .
Solution
Let the dimensions of theprinted region be x and y, Pthe printed area, and A thepaper area. We wish tomaximize P = xy subject tothe constraint that
A = (x+ 2)(y+ 3) ≡ 120
Isolating y in A ≡ 120 gives
y =120x+ 2
− 3 which yields
P = x(
120x+ 2
− 3)
=120xx+ 2
−3x
The domain of P is (0,∞)
.
.Lorem ipsum dolor sit amet,consectetur adipiscing elit. Namdapibus vehicula mollis. Proin nectristique mi. Pellentesque quisplacerat dolor. Praesent a nisl diam.Phasellus ut elit eu ligula accumsaneuismod. Nunc condimentumlacinia risus a sodales. Morbi nuncrisus, tincidunt in tristique sit amet,ultrices eu eros. Proin pellentesquealiquam nibh ut lobortis. Ut etsollicitudin ipsum. Proin gravidaligula eget odio molestie rhoncussed nec massa. In ante lorem,imperdiet eget tincidunt at, pharetrasit amet felis. Nunc nisi velit,tempus ac suscipit quis, blanditvitae mauris. Vestibulum ante ipsumprimis in faucibus orci luctus etultrices posuere cubilia Curae;
.1.5 cm
.1.5 cm
.1cm
.1cm
.x
.y
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 29 / 31
. . . . . .
Solution (Concluded)
We want to find the absolute maximum value of P. Taking derivatives,
dPdx
=(x+ 2)(120)− (120x)(1)
(x+ 2)2− 3 =
240− 3(x+ 2)2
(x+ 2)2
There is a single critical point when
(x+ 2)2 = 80 =⇒ x = 4√5− 2
(the negative critical point doesn’t count). The second derivative is
d2Pdx2
=−480
(x+ 2)3
which is negative all along the domain of P. Hence the unique criticalpoint x =
(4√5− 2
)cm is the absolute maximum of P. This means
the paper width is 4√5 cm, and the paper length is
1204√5= 6
√5 cm.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 30 / 31
. . . . . .
Summary
I Remember the checklistI Ask yourself: what is the
objective?I Remember your geometry:
I similar trianglesI right trianglesI trigonometric functions
Name [_
Problem Solving StrategyDraw a Picture
Kathy had a box of 8 crayons.She gave some crayons away.She has 5 left.How many crayons did Kathy give away?
UNDERSTAND•
What do you want to find out?Draw a line under the question.
You can draw a pictureto solve the problem.
crayons
What number do Iadd to 5 to get 8?
8 - = 55 + 3 = 8
CHECK
Does your answer make sense?Explain.
Draw a picture to solve the problem.Write how many were given away.I. I had 10 pencils.
I gave some away.I have 3 left. How manypencils did I give away?
~7
What numberdo I add to 3to make 10?
13iftill:ii ?11
ftI'•'«II
ft A
H 11M i l
U U U U> U U
I I
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 31 / 31