Question

Three of the sides will require fencing and the fourth wall already exists. If the farmer has 116 feet of fencing, what are the dimensions of the region with the largest area

Answer 1

Answer:

29 ft x 58 ft

Step-by-step explanation:

Let x be the length of each side perpendicular to the wall, and y be the length of the side parallel to the wall.

The amount of wire available is:

$116 = 2x+y\\y=116-2x$

The area of the region is:

$A=xy=x(116-2x)\\A(x)=116x-2x^2$

The value of 'x' for which the derivate of the area function is zero will yield the maximum area:

$A(x)=116x-2x^2\\A'(x) = 116-4x=0\\x=29\ ft$

The value of y is:

$y=116-2*29\\y=58\ ft$

The dimensions of the region with the largest area are 29 ft x 58 ft.