find the dimensions of the rectangle that will enclose the most area, and show steps please.
A fence is to be built to enclose a rectangular area. The fence along three sides is to be made of material
that costs $5 per foot. The material for the fourth side costs $15 per foot.
(B) If $3,000 is available for the fencing, find the dimensions of the rectangle that will enclose the most area.
We need to find the maximum area that can be enclosed in the rectangular area with a cost of 3000.
Let the sides be x and y.
Let the three sides that will cost 5 dollars per foot are x, y, and y and for the fourth side y will cost 15 dollars per foot.
The cost for the three sides is 5*(x+y+y)= 5(x+2y)
The cost for the forth side is 15*(x).
The total cost is 3000
==> 5(x+2y) + 15x = 3000
==> 5x + 10y + 15x = 3000
==> 20x + 10y= 3000
==> 2x + y = 300
Let the area of the rectangle be A.
==> A= xy
But we know that 2x+y = 300 ==> y= 300-2x
==> A = x(300-2x)
==> A = 300x - 2x^2
Now we need to find the maximum area which is the zero of the derivative A'.
==> A'= 300-4x = 0
==> 4x = 300
==> x = 300/4 = 75.
==> y= 300-2x = 300-150 = 150
Then, the dimensions for a maximum area are x=75 ft and y=150 ft.