Problem of the Day No calculator!

If f(x) = ln x for 0 < x < 2x2 ln 2 for 2 < x < 4

{then lim f(x) is x⇒2

A) ln 2 C) ln 16 E) nonexistent B) ln 8 D) 4

Problem of the Day

If f(x) = ln x for 0 < x < 2x2 ln 2 for 2 < x < 4

{then lim f(x) is x⇒2

A) ln 2 C) ln 16 E) nonexistent B) ln 8 D) 4

limit from left does not equal limit from right at 2

Optimization ProblemsOne of the most common uses

of calculus is to determine minimum and maximum values. Some examples aregreatest profit

least costleast timegreatest voltageoptimum sizegreatest strengthgreatest distancegreatest volumeleast materialleast size

A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?

Method 1 - Guess and check

surface area = area of base + 4 (area of a side)108 = (x)(x) + 4(x)

(h)

xxh

108 = x2 + 4xh

A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?

Method 1 - Guess and check

xxh

108 = x2 + 4xhvolume = x2h

A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?

Method 1 - Guess and check

xxh 108 = x2 +

4xhlength of side

volume345678

volume = x2hheight

A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?

Method 1 - Guess and check

xxh 108 = x2 +

4xh

length of side height

volume345678

74.2592103 ¾108103¼88

8.255.754.1532.111.375

volume = x2h

Method 1 - Guess and check

xxh 108 = x2 +

4xhlength of side height

volume345678

74.2592103 ¾108103¼88

8.255.754.1532.111.375

Any problems with this method?

volume = x2h

A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?

Method 2 - Use Calculus

xxh

108 = x2 + 4xh

surface area = area of base + 4 (area of a side)

volume = x2h

What is to be maximized?

What is this in Calculus?

A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?

Method 2 - Use Calculus

xxh

108 = x2 + 4xh

surface area = area of base + 4 (area of a side)

volume = x2h

What is to be maximized? VolumeWhat is this in Calculus? Critical Number

And Critical Numbers come from?

A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?

Method 2 - Use Calculus

xxh

108 = x2 + 4xh

surface area = area of base + 4 (area of a side)

volume = x2h

What is to be maximized? VolumeWhat is this in Calculus? Critical NumberAnd Critical Numbers come from? First Derivative Set Equal to Zero

A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?

Method 2 - Use Calculus

xxh

108 = x2 + 4xh

surface area = area of base + 4 (area of a side)

volume = x2h

How can we eliminate a variable?

A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?

Method 2 - Use Calculus

xxh

108 = x2 + 4xh

surface area = area of base + 4 (area of a side)

volume = x2h

How can we eliminate a variable? Use the equation not being maximized to solve for one of the variables and substitute the result into the equation being maximized.

Procedure to Maximizing or Minimizing1) Develop 2 equations2) Use equation not being maximized to eliminate a variable3) Determine feasible domain for x4) Differentiate5) Find critical numbers6) Evaluate at endpoints of domain and at critical numbers7) Pick value that gives the largest or smallest value

1) Develop 2 equations108 = x2 +

4xhvolume = x2h

2) Use equation not being maximized to eliminate a variable108 = x2 +

4xhh = 108 - x2

4x

V = x2 hV= (x2 ) 108 - x2

4xV = 27x - ¼x3

3) Determine feasible domain for x1) x must be greater than zero

2) The largest x would be when the height is zero. Then 108 = x2 + 0 and x2 = √108

Thus 0 < x < √1084) DifferentiateV = 27x -

¼x3

dV = 27 - ¾x2

dx

dV = 27 - ¾x2

dx

0 = 27 - ¾x2

3x2 = 108 x = ±6

5) Find critical numbers

6) Evaluate at endpoints of domain and at critical numbersf(0) = 0

f(6) = 108f(√108) = 0

f(0) = 0f(6) = 108f(√108) = 0

7) Pick value that gives the largest or smallest value

maximum

so x = 6 and h = 3

dimensions are 6 x 6 x 3

Find 2 positive integers whose product is 192 and the sum of the first plus three times the second is a minimum.

Find 2 positive integers whose product is 192 and the sum of the first plus three times the second is a minimum.1) Develop 2 equations

2) Use equation not being maximized to eliminate a variable

Find 2 positive integers whose product is 192 and the sum of the first plus three times the second is a minimum.3) Determine feasible domain for x

4) Differentiate

5) Find critical numbers

Find 2 positive integers whose product is 192 and the sum of the first plus three times the second is a minimum.6) Evaluate at endpoints of domain and at critical numbers

7) Pick value that gives the largest or smallest value

Find 2 positive integers whose product is 192 and the sum of the first plus three times the second is a minimum.

The numbers are 24 and 8

Find the length and width of a rectangle that has an area of 64 and a minimum perimeter.

Find the length and width of a rectangle that has an area of 64 and a minimum perimeter.

Area = l . w64 = l . w64/w = l

Perimeter = 2l + 2wPerimeter = 2(64/w) + 2w

y' = -128/w2 + 20 = -128/w2 + 28 = wLength = 8 and

width = 8

Find the point on the graph of the function f(x) = x2 that is closest to the point (2, ½).

Find the point on the graph of the function f(x) = x2 that is closest to the point (2, ½).

y = x2

distance = √(2 - x)2 + (½ - y)2

distance = (x4 - 4x + 4.25)½

distance ' = ½(x4 - 4x + 4.25)-½ (4x3 - 4) 0 = ½(x4 - 4x + 4.25)-½ (4x3 -

4)1 = x

A tank with a rectangular base and rectangular sides is to be open at the top. It is to be constructed so that its width is 4 meters and its volume is 36 cubic meters. If building the tank costs $10 per square meter for the base and $5 per square meter for the sides, what is the cost of the least expensive tank?

A tank with a rectangular base and rectangular sides is to be open at the top. It is to be constructed so that its width is 4 meters and its volume is 36 cubic meters. If building the tank costs $10 per square meter for the base and $5 per square meter for the sides, what is the cost of the least expensive tank?

$330

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