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:
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) [tex]C=A_{base}. C_{base} +A_{top}. C_{top} +4.A_{sides}. C_{sides}[/tex]
Replacing square areas for base and top and rectangular area for sides:
[tex]C=x^{2} . C_{base} +x^{2} . C_{top} +4.x.l. C_{sides}[/tex]
[tex]C=x^{2}.30 \frac{cents}{ft^{2} } +x^{2} . 20 \frac{cents}{ft^{2} }+4.x.l. 10 \frac{cents}{ft^{2} }[/tex]
(3) [tex]C=50x^{2}\frac{cents}{ft^{2} }+40.x.l\frac{cents}{ft^{2} }[/tex]
Let´s clear l from equation (1) just to replace it in equation (3):
/[tex]20 ft^{3}=x^{2}.l[/tex]
(4) [tex]l=\frac{20 ft^{3}}{x^{2}}[/tex]
Replacing l found in (3):
[tex]C=50x^{2}\frac{cents}{ft^{2} }+40.x.\frac{20 ft^{3}}{x^{2}}\frac{cents}{ft^{2} }[/tex]
(5) [tex]C=50x^{2}\frac{cents}{ft^{2} }+\frac{800}{x}ft.cents[/tex]
To determine an equation minimum is necessary to equal its derivative to zero. So, we have to find equation (5) derivative ([tex]\frac{dC}{dx}[/tex]):
[tex]\frac{dC}{dx} =50.(2.x)\frac{cents}{ft^{2} }+800.(-\frac{1}{x^{2}})ft.cents=0[/tex]ft.cents=0[/tex]
[tex]\frac{dC}{dx} =100x\frac{cents}{ft^{2} }-800.\frac{1}{x^{2}}ft.cents=0[/tex]
[tex]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[/tex]
Finally with x value found above, we can obtain l from equation (4):
[tex]l=\frac{20 ft^{3}}{x^{2}}[/tex]
[tex]l=\frac{20 ft^{3}}{(2ft)^{2}}[/tex]
[tex]l=\frac{20 ft^{3}}{4ft^{2}}\\l=5ft[/tex]
So, dimensions for the minimum cost are:
x = 2 feet
l = 5 feet
