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.

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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.**

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