A box with a square base and open top must have a volume of 62500 cm³. 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 x, the length of one side of the square base. [Hint: use the volume formula to express the height of the box in terms of x.] Simplify your formula as much as possible. A(x) = Next, find the derivative, A'(x). A'(x) = Now, calculate when the derivative equals zero, that is, when A'(x) = 0. [Hint: multiply both sides by ² .] A'(x) = 0 when x = We next have to make sure that this value of a gives a minimum value for the surface area. Let's use the second derivative test. Find A"(x). A"(x) = Evaluate A"(x) at the x-value you gave above.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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A box with a square base and open top must have a volume of 62500 cm³. 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 x, the length of one side of the square
base.
[Hint: use the volume formula to express the height of the box in terms of x.]
Simplify your formula as much as possible.
A(x) =
Next, find the derivative, A'(x).
A'(x)
=
Now, calculate when the derivative equals zero, that is, when A'(x) = 0. [Hint: multiply both sides by ²
.]
A'(x) = 0 when x =
We next have to make sure that this value of a gives a minimum value for the surface area. Let's use the
second derivative test. Find A"(x).
A"(x) =
Evaluate A"(x) at the x-value you gave above.
NOTE: Since your last answer is positive, this means that the graph of A(x) is concave up around that
value, so the zero of A'(x) must indicate a local minimum for A(x). (Your boss is happy now.)
Transcribed Image Text:A box with a square base and open top must have a volume of 62500 cm³. 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 x, the length of one side of the square base. [Hint: use the volume formula to express the height of the box in terms of x.] Simplify your formula as much as possible. A(x) = Next, find the derivative, A'(x). A'(x) = Now, calculate when the derivative equals zero, that is, when A'(x) = 0. [Hint: multiply both sides by ² .] A'(x) = 0 when x = We next have to make sure that this value of a gives a minimum value for the surface area. Let's use the second derivative test. Find A"(x). A"(x) = Evaluate A"(x) at the x-value you gave above. NOTE: Since your last answer is positive, this means that the graph of A(x) is concave up around that value, so the zero of A'(x) must indicate a local minimum for A(x). (Your boss is happy now.)
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