A farmer wants to build a rectangular pen with 80 feet of fencing. The pen will be built against the wall of the barn, so one side of the rectangle won’t need a fence. What dimensions will maximize the area of the pen?

Let x represent the length of fence in the direction parallel to the wall. Then the other dimension of the rectangular pen is (80 -x)/2. The area is the product of these dimensions:

area = x(80 -x)/2

This function describes a downward-opening parabola with zeros at x=0 and x=80. The vertex (maximum) is halfway between the zeros, at x=40.

The dimensions are 40 ft parallel to the wall and 20 ft out from the wall.

ashishdwivedilVT ashishdwivedilVT · Answer 2 · 2022-03-07T13:08:33+01:00

Length l=40 and Breadth b=20 will maximize the area of the pen.

let us take 'l' as the length of the rectangular pen.

'b' as the width of the rectangular pen.

let us assume that the barn will be built opposite to length.

so, according to the given condition

l +b+b=80

l+2b=80......(1)

area of the rectangular pen = lb= (80-2b)b

f(b)= (80-2b)b.......(2)

How to check the local maxima?

to get local maxima, differentiate the function and equate to zero, get the point say it 'x'

again check double derivative if its value is negative the point 'x' will give the maximum value of the function.

to maximize the area

let us derivate the f(b)= (80-2b)b

f'(b)= 80-4b=0

b=20

f"(b)= -4(-ve)

means we will have local maxima at b=20

it means at b=20, we will get maximum area.

l = 80-2b=80-2*20=40

therefore, l=40 and b=20 will maximize the area of the pen.

to get more about maxima and minima refer to the link,

https://brainly.com/question/6422517