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Artyom0805 [142]
3 years ago
11

A rectangular box is to have a square base and a volume of 12 ft3. If the material for the base costs $0.17/ft2, the material fo

r the sides costs $0.10/ft2, and the material for the top costs $0.13/ft2, determine the dimensions (in ft) of the box that can be constructed at minimum cost.
a.length ft

b.width ft

c.height ft
Mathematics
1 answer:
katen-ka-za [31]3 years ago
7 0

Answer:

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet

Step-by-step explanation:

Let the dimensions of the box be x, y and z

The rectangular box has a square base.

Therefore, Volume of the boxV=x^2z

Volume of the box=12 ft^3\\

Therefore, x^2z=12\\z=\frac{12}{x^2}

The material for the base costs \$0.17/ft^2, the material for the sides costs \$0.10/ft^2, and the material for the top costs \$0.13/ft^2.

Area of the base =x^2

Cost of the Base =\$0.17x^2

Area of the sides =4xz

Cost of the sides==\$0.10(4xz)

Area of the Top =x^2

Cost of the Base =\$0.13x^2

Total Cost, C(x,z) =0.17x^2+0.13x^2+0.10(4xz)

Substituting z=\frac{12}{x^2}

C(x) =0.17x^2+0.13x^2+0.10(4x)(\frac{12}{x^2})\\C(x)=0.3x^2+\frac{4.8}{x} \\C(x)=\dfrac{0.3x^3+4.8}{x}

To minimize C(x), we solve for the derivative and obtain its critical point

C'(x)=\dfrac{0.6x^3-4.8}{x^2}\\Setting \:C'(x)=0\\0.6x^3-4.8=0\\0.6x^3=4.8\\x^3=4.8\div 0.6\\x^3=8\\x=\sqrt[3]{8}=2

Recall: z=\frac{12}{x^2}=\frac{12}{2^2}=3\\

Therefore, the dimensions that minimizes the cost of the box are:

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet

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