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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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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Answer 1 · 2015-03-24T21:32:54

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

When these are folded up they give a box with a height of $x$ $x$ feet
and a base of $20 - 2 x$ $20 - 2x$ feet by $20 - 2 x$ $20-2x$ feet
for a volume
$V = x {(20 - 2 x)}^{2} = 400 x - 80 x^{2} + 4 x^{3}$ $V = x(20-2x)^2= 400x -80x^2 + 4x^3$

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

$\frac{d V}{d x} = 400 - 160 x + 12 x^{2}$ $(dV)/(dx) = 400-160x+12x^2$

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

Since $x = 10$ $x=10$ gives a Volume of $0$ $0$

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

The resulting box will be
$3 \frac{1}{3} \times 13 \frac{1}{3} \times 13 \frac{1}{3}$ $3 1/3 xx 13 1/3 xx 13 1/3$ feet

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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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