glyperezrl
2022-07-16
Answered

What will the dimensions of the resulting cardboard box be if the company wants to maximize the volume and they start with a flat piece of square cardboard 20 feet per side, and then cut smaller squares out of each corner and fold up the sides to create the box?

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coolng90qo

Answered 2022-07-17
Author has **14** answers

Suppose that the squares removed from each corner are x feet by x feet each.

When these are folded up they give a box with a height of x feetand a base of 20−2x feet by 20−2x feet for a volume

$V=x{(20-2x)}^{2}=400x-80{x}^{2}+4{x}^{3}$

To find the critical point(s) take the derivative of V, set it to zero, and solve for x.

$\frac{dV}{dx}=400-160x+12{x}^{2}$

$=4(3x-10)(x-10)=0$

Since x=10 gives a Volume of 0

so the critical point for the Volume that is it's maximum occurs when $x=\frac{10}{3}$

The resulting box will be

$3\frac{1}{3}\times 13\frac{1}{3}\times 13\frac{1}{3}$

When these are folded up they give a box with a height of x feetand a base of 20−2x feet by 20−2x feet for a volume

$V=x{(20-2x)}^{2}=400x-80{x}^{2}+4{x}^{3}$

To find the critical point(s) take the derivative of V, set it to zero, and solve for x.

$\frac{dV}{dx}=400-160x+12{x}^{2}$

$=4(3x-10)(x-10)=0$

Since x=10 gives a Volume of 0

so the critical point for the Volume that is it's maximum occurs when $x=\frac{10}{3}$

The resulting box will be

$3\frac{1}{3}\times 13\frac{1}{3}\times 13\frac{1}{3}$

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