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ki77a [65]
4 years ago
15

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. T

he 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.

Mathematics
1 answer:
sashaice [31]4 years ago
7 0

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]

<u>You may find attached a draw with the dimentions of the box.</u>

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=0ft.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

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