What are the dimensions of a box that will use the minimum amount of materials, if the firm needs a closed box in which the bottom is in the shape of a rectangle, where the length being twice as long as the width and the box must hold 9000 cubic inches of material?

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What are the dimensions of a box that will use the minimum amount of materials, if the firm needs a closed box in which the bottom is in the shape of a rectangle, where the length being twice as long as the width and the box must hold 9000 cubic inches of material?

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Answer 1 · 2015-02-24T11:16:51

Let's begin by putting in some definitions.

If we call $h$ $h$ the height of the box and $x$ $x$ the smaller sides (so the larger sides are $2 x$ $2x$ , we can say that volume

$V = 2 x \cdot x \cdot h = 2 x^{2} \cdot h = 9000$ $V=2x*x*h=2x^2*h=9000$ from which we extract $h$ $h$

$h = \frac{9000}{2 x^{2}} = \frac{4500}{x^{2}}$ $h=9000/(2x^2)=4500/x^2$

Now for the surfaces (=material)

Top&bottom: $2 x \cdot x$ $2x*x$ times $2 \to$ $2->$ Area= $4 x^{2}$ $4x^2$
Short sides: $x \cdot h$ $x*h$ times $2 \to$ $2->$ Area= $2 x h$ $2xh$
Long sides: $2 x \cdot h$ $2x*h$ times $2 \to$ $2->$ Area= $4 x h$ $4xh$

Total area:

$A = 4 x^{2} + 6 x h$ $A=4x^2+6xh$

Substituting for $h$ $h$

$A = 4 x^{2} + 6 x \cdot \frac{4500}{x^{2}} = 4 x^{2} + \frac{27000}{x} = 4 x^{2} + 27000 x^{- 1}$ $A=4x^2+6x*4500/x^2=4x^2+27000/x=4x^2+27000x^-1$

To find the minimum, we differentiate and set $A'$ to $0$

$A'=8x-27000x^-2=8x-27000/x^2=0$

Which leads to $8x^3=27000->x^3=3375->x=15$

Answer :
Short side is $15$ inches
Long side is $2*15=30$ inches
Height is $4500/15^2=20$ inches

Check your answer! $15*30*20=9000$

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What are the dimensions of a box that will use the minimum amount of materials, if the firm needs a closed box in which the bottom is in the shape of a rectangle, where the length being twice as long as the width and the box must hold 9000 cubic inches of material?

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