A box with a square base and open top must have a volume of 256 cm. 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. 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. %3D A'(x) = 0 when x = We next have to make sure that this value of x 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).

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Author:Bruce Crauder, Benny Evans, Alan Noell
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A box with a square base and open top must have a volume of 256 cm°. 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.
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.
A'(x) = 0 when x =
We next have to make sure that this value of x 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).
Transcribed Image Text:A box with a square base and open top must have a volume of 256 cm°. 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. 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. A'(x) = 0 when x = We next have to make sure that this value of x 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).
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