A box with a square base and open top must have a volume of 37044 cm3cm3. We wish to find the dimensions of the box that minimize the amount of material used.First, find a formula for the surface area of the box in terms of only xx, the length of one side of the square base.[Hint: use the volume formula to express the height of the box in terms of xx.]Simplify your formula as much as possible.A(x)=A(x)=   Next, find the derivative, A'(x)A′(x).A'(x)=A′(x)=   Now, calculate when the derivative equals zero, that is, when A'(x)=0A′(x)=0. [Hint: multiply both sides by x2x2.]A'(x)=0A′(x)=0 when x=x=We next have to make sure that this value of xx gives a minimum value for the surface area. Let's use the second derivative test. Find A"(x)(x).A"(x)=(x)=   Evaluate A"(x)(x) at the xx-value you gave above.

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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A box with a square base and open top must have a volume of 37044 cm3cm3. We wish to find the dimensions of the box that minimize the amount of material used.

First, find a formula for the surface area of the box in terms of only xx, the length of one side of the square base.
[Hint: use the volume formula to express the height of the box in terms of xx.]
Simplify your formula as much as possible.
A(x)=A(x)=   

Next, find the derivative, A'(x)A′(x).
A'(x)=A′(x)=   

Now, calculate when the derivative equals zero, that is, when A'(x)=0A′(x)=0. [Hint: multiply both sides by x2x2.]
A'(x)=0A′(x)=0 when x=x=

We next have to make sure that this value of xx gives a minimum value for the surface area. Let's use the second derivative test. Find A"(x)(x).
A"(x)=(x)=   

Evaluate A"(x)(x) at the xx-value you gave above.

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