Bill wants to fence three sides of a rectangular exercise ya

Bill wants to fence three sides of a rectangular exercise yard for his dog. The fourth side of the exercise yard will be a side of the house. He has 160 feet of fencing available. Find the dimensions that will enclose the maximum area.

The fence parallel to the house is ______ feet, the fence perpendicular to the house is ____________ feet and the area of the yard is ___________________square feet.

Solution

Dear Student

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As per the problem statement let us consider the fence side parallel to house be \'l\' and the side perpendicular to the house be \'b\'

Hence l + 2b=160 (length of fence)

l=160-2b

Now area of the exercise yard will be

A=l X b

= (160-2b)b

= 160b-2b^2

Now for maximizing let us differentiate the above equation w.r.t. b which gives

160-4b=0                     Eq(1)

b=40 feet.

& Hence l=160-2b

l=80 feet

Double differentiating eq(1) gives -4 which is negative and hence proves that value if calculated \'b\' shall correspond to maximum area.

Maximum area = l X b

= 80 X 40

= 3200 feet^2

The fence parallel to the house is __80 ___ feet, the fence perpendicular to the house is ____40 _______ feet and the area of the yard is _______3200____________square feet.

Solved !!