Question

(1 point)
a rectangular storage container with an open top is to have a volume of $96m^{3}$. the length of its base is twice the width. material for the base costs $36 per square meter. material for the sides costs $32 per square meter. find the dimensions and the cost of materials for the cheapest such container.

• width:
• length:
• height:
• cost: $

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Explanation:

Step1: Define variables

Let the width of the base be \(x\) meters. Then the length of the base is \(2x\) meters. Let the height be \(h\) meters.
The volume \(V = l\times w\times h\), so \(96=(2x)\times x\times h\), which gives \(h=\frac{96}{2x^{2}}=\frac{48}{x^{2}}\).

Step2: Write the cost function

The area of the base \(A_{base}=2x\times x = 2x^{2}\). The cost of the base \(C_{base}=36\times2x^{2}=72x^{2}\).
The area of the sides: There are two sides with dimensions \(x\times h\) and two sides with dimensions \(2x\times h\). So \(A_{sides}=2(xh + 2xh)=6xh\). The cost of the sides \(C_{sides}=32\times6xh = 192xh\).
The total cost \(C(x)=72x^{2}+192x\times\frac{48}{x^{2}}=72x^{2}+\frac{9216}{x}\).

Step3: Find the derivative of the cost function

Using the power rule, \(C^\prime(x)=144x-\frac{9216}{x^{2}}\).

Step4: Set the derivative equal to zero and solve for \(x\)

\(144x-\frac{9216}{x^{2}} = 0\)
Multiply through by \(x^{2}\): \(144x^{3}-9216 = 0\)
\(x^{3}=\frac{9216}{144}=64\)
\(x = 4\)

Step5: Find the other dimensions

When \(x = 4\), \(l=2x = 8\)
\(h=\frac{48}{x^{2}}=\frac{48}{16}=3\)

Step6: Calculate the cost

\(C(4)=72\times(4)^{2}+\frac{9216}{4}=72\times16 + 2304=1152+2304=3456\)

Answer:

Width: \(4\)m
Length: \(8\)m
Height: \(3\)m
Cost: \(\$3456\)