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An open box is to be constructed so that the length of the base is $3$ times larger than the width of the base. If the cost to construct the base is $5$ dollars per square foot and the cost to construct the four sides is $4$ dollars per square foot, determine the dimensions for a box to have volume $= 89$ cubic feet which would minimize the cost of construction.

Height Value of dimensions

So $L = 4 W$, $V w/h 89$, $W^2 H = 89/4$, or $H = 89/4 /w^2$

I'm so confused what to do next If someone could help. Thanks.

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1 Answer

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You're told $L = 3 W$, so we'll make this substitution whenever we can.

The area of the base is $LW$, so the cost of the base is $5LW = 15 W^2$.

The sides have area $LH$ or $WH$, so the total cost of the sides is

$$4(2LH + 2WH) = 4(6WH + 2WH) = 32WH$$

and then the total cost of the box is $32WH + 15W^2$.

You know the total volume is $89$, so

$$LWH = 89,\\ 3W^2H = 89,\\ H = \dfrac{89}{3W^2}$$

and so total cost is

$$ C = 32W\left(\dfrac{89}{3W^2} \right) + 15W^2. $$

Now simplify this expression and find the minimum (i.e. the width that makes cost a minimum. Then work back to find $L$ and $H$ from the relationships above)

Explicitly, $C = \dfrac{2848}{3W} + 15W^2$ and so $$ \frac{dC}{dW} = -\dfrac{2848}{3W^2} + 30W. $$

Set this equal to $0$:

$$ 0 = -\dfrac{2848}{3W^2} + 30W $$

multiply by $W^2$: $$ 0 = -\dfrac{2848}{3} + 30W^3 \\ \dfrac{2848}{3} = 30W^3 \\ \dfrac{2848}{90} = W^3 \\ W = \sqrt[3]{\dfrac{2848}{90}} \simeq 3.16 $$

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