Question

rectangular box with a square base is to have a volume of 20 cubic feet. The material for the base costs 30 cents/square foot. The material for the sides costs 10 cents/square foot. The material for the top costs 20 cents/square foot. Determine the dimensions that will yield minimum cost. Let x = length of the side of the base.

Answer 1

Answer:

The dimensions for the minimum cost are:

x = 2 feet

l = 5 feet (l is the hight of the box)

Step-by-step explanation:

Let´s find the data given:

• rectangular box
• square base
• volumen $V = 20 ft^{3}$

• base material cost $Cost_{base} = 30 \frac{cents}{ft^{2} }$

• sides material cost $Cost_{sides} = 10 \frac{cents}{ft^{2} }$

• top material cost $Cost_{top} = 20 \frac{cents}{ft^{2} }$[/tex]

You may find attached a draw with the dimentions of the box.

To know the box volume (V) we have to multiply base sides (x) and the box hight (l) as shown below:

(1)   V = x . x . l

And to determine the total cost, it´s necessary to add base, top and the 4 sides costs. Each one of then can be calculated as the material cost multiplied by the area. Then, total cost C is:

(2)  $C=A_{base}. C_{base} +A_{top}. C_{top} +4.A_{sides}. C_{sides}$

Replacing square areas for base and top and rectangular area for sides:

$C=x^{2} . C_{base} +x^{2} . C_{top} +4.x.l. C_{sides}$

$C=x^{2}.30 \frac{cents}{ft^{2} } +x^{2} . 20 \frac{cents}{ft^{2} }+4.x.l. 10 \frac{cents}{ft^{2} }$

(3)   $C=50x^{2}\frac{cents}{ft^{2} }+40.x.l\frac{cents}{ft^{2} }$

Let´s clear l from equation (1) just to replace it in equation (3):

/$20 ft^{3}=x^{2}.l$

(4)  $l=\frac{20 ft^{3}}{x^{2}}$

Replacing l found in (3):

$C=50x^{2}\frac{cents}{ft^{2} }+40.x.\frac{20 ft^{3}}{x^{2}}\frac{cents}{ft^{2} }$

(5)  $C=50x^{2}\frac{cents}{ft^{2} }+\frac{800}{x}ft.cents$

To determine an equation minimum is necessary to equal its derivative to zero. So, we have to find equation (5) derivative ($\frac{dC}{dx}$):

$\frac{dC}{dx} =50.(2.x)\frac{cents}{ft^{2} }+800.(-\frac{1}{x^{2}})ft.cents=0$ft.cents=0[/tex]

$\frac{dC}{dx} =100x\frac{cents}{ft^{2} }-800.\frac{1}{x^{2}}ft.cents=0$

$100x\frac{cents}{ft^{2} }=800.\frac{1}{x^{2}}ft.cents\\x^{3}=\frac{800ft.cents}{100\frac{cents}{ft^{2} }} \\x^{3}=8ft^{3} \\x=2ft$

Finally with x value found above, we can obtain l from equation (4):

$l=\frac{20 ft^{3}}{x^{2}}$

$l=\frac{20 ft^{3}}{(2ft)^{2}}$

$l=\frac{20 ft^{3}}{4ft^{2}}\\l=5ft$

So, dimensions for the minimum cost are:

x = 2 feet

l = 5 feet