Question

(1 point) a rectangular storage container with an open top is to have a volume of 12 cubic meters. the length of its base is twice the width. material for the base costs 10 dollars per square meter. material for the sides costs 4 dollars per square meter. find the cost of materials for the cheapest such container. total cost = (round to the nearest penny and include monetary units. for example, if your answer is 1.095, enter $1.10 including the dollar sign and second decimal place.) video examples: solving similar problems

Explanation:

Step1: Let the width of the base be $w$ meters. Then the length of the base is $2w$ meters. Let the height of the container be $h$ meters.

The volume $V = l\times w\times h$. Given $V = 12$ cubic - meters and $l = 2w$, so $12=2w\times w\times h=2w^{2}h$, and $h=\frac{6}{w^{2}}$.

Step2: Calculate the cost function.

The area of the base $A_{base}=l\times w = 2w\times w = 2w^{2}$ square - meters. The cost of the base $C_{base}=10\times2w^{2}=20w^{2}$ dollars.
The area of the four sides: Two sides have area $l\times h = 2w\times h$ and two sides have area $w\times h$. The total area of the sides $A_{sides}=2(2w\times h)+2(w\times h)=6wh$ square - meters. The cost of the sides $C_{sides}=4\times6wh = 24wh$ dollars.
The total cost function $C(w)=20w^{2}+24wh$. Substitute $h = \frac{6}{w^{2}}$ into the cost function: $C(w)=20w^{2}+24w\times\frac{6}{w^{2}}=20w^{2}+\frac{144}{w}$.

Step3: Find the derivative of the cost function.

Differentiate $C(w)$ with respect to $w$. $C^\prime(w)=40w-\frac{144}{w^{2}}$.

Step4: Set the derivative equal to zero to find critical points.

$40w-\frac{144}{w^{2}} = 0$. Multiply through by $w^{2}$ to get $40w^{3}-144 = 0$. Then $w^{3}=\frac{144}{40}=\frac{18}{5}$, and $w=\sqrt[3]{\frac{18}{5}}\approx1.53$ meters.

Step5: Find the second - derivative of the cost function to confirm it's a minimum.

Differentiate $C^\prime(w)$ with respect to $w$: $C^{\prime\prime}(w)=40+\frac{288}{w^{3}}$. When $w=\sqrt[3]{\frac{18}{5}}$, $C^{\prime\prime}(w)>0$, so it's a minimum.

Step6: Calculate the minimum cost.

Substitute $w=\sqrt[3]{\frac{18}{5}}$ into the cost function $C(w)$.
$C(\sqrt[3]{\frac{18}{5}})=20(\sqrt[3]{\frac{18}{5}})^{2}+\frac{144}{\sqrt[3]{\frac{18}{5}}}$.
$(\sqrt[3]{\frac{18}{5}})^{2}=\sqrt[3]{\frac{324}{25}}$ and $\frac{1}{\sqrt[3]{\frac{18}{5}}}=\sqrt[3]{\frac{5}{18}}$.
$C(\sqrt[3]{\frac{18}{5}})=20\sqrt[3]{\frac{324}{25}} + 144\sqrt[3]{\frac{5}{18}}$.
$20\sqrt[3]{\frac{324}{25}}\approx20\times2.34=46.8$ and $144\sqrt[3]{\frac{5}{18}}\approx144\times0.65 = 93.6$.
$C(\sqrt[3]{\frac{18}{5}})\approx46.8 + 93.6=140.40$ dollars.

Answer:

$\$140.40$