Question

a storage shed is to be built in the shape of a box with a square base. it is to have a volume of 125 cubic feet. the concrete for the base costs $6 per square foot, the material for the roof costs $9 per square foot, and the material for the sides costs $7.50 per square foot. find the dimensions of the most economical shed. the length of one side of the sheds base is □ ft. the height of the shed is □ ft.

Explanation:

Step1: Let the side length of the square base be \(x\) feet and the height be \(h\) feet.

The volume \(V=x^{2}h\), and since \(V = 125\), we have \(h=\frac{125}{x^{2}}\).

Step2: Calculate the cost function \(C\).

The area of the base is \(x^{2}\), and its cost is \(6x^{2}\). The area of the four sides is \(4xh\), and its cost is \(7.5\times4xh = 30xh\). The area of the roof (two congruent rectangles with area \(x\times\frac{\sqrt{x^{2}/4 + h^{2}}}{1}\) each, for a square - based shed with a simple roof design, assuming a gable - type roof with the slant height related to \(x\) and \(h\); if we assume a flat - like roof for simplicity of cost calculation, the area of the roof is \(x^{2}\) and its cost is \(9x^{2}\)). So \(C=6x^{2}+30xh + 9x^{2}=15x^{2}+30xh\).
Substitute \(h=\frac{125}{x^{2}}\) into the cost function: \(C(x)=15x^{2}+30x\times\frac{125}{x^{2}}=15x^{2}+\frac{3750}{x}\).

Step3: Find the derivative of \(C(x)\).

Using the power rule, \(C^\prime(x)=30x-\frac{3750}{x^{2}}\).

Step4: Set \(C^\prime(x) = 0\) to find critical points.

\(30x-\frac{3750}{x^{2}}=0\).
Multiply through by \(x^{2}\): \(30x^{3}-3750 = 0\).
\(x^{3}=\frac{3750}{30}=125\).
So \(x = 5\) (since \(x>0\)).

Step5: Find \(h\).

Substitute \(x = 5\) into \(h=\frac{125}{x^{2}}\), we get \(h=\frac{125}{25}=5\).

Answer:

The length of one side of the shed's base is \(5\) ft. The height of the shed is \(5\) ft.