Question

One side of a rectangular field is adjacent to a river. There are 160 yards of fencing available for the other three sides. What dimensions of the field will result in the maximum area?

Answer 1

The dimensions of 40 yards by 80 yards of the field will result in maximum area of 3200 yards squared.

Explanation:

Let "x" and "y" be the lengths of the sides of rectangle.

So, the perimeters of three sides is:

2x + y = 160 yards

y = 160 - 2x ______ eqn (1)

Area = A = xy = x(160 - 2x)

A(x) = 160x - 2x² [in terms of x]

Taking 1st and setting value to zero:

A'(x) = 0 = 160 - 4x

4x = 160

x = 40 yards

Now, taking 2nd derivative to check whether value is minimum or maximum.

A"(x) = -4

since, A"(x) < 0

therefore, it means the value x = 40 yards is maximum.

Thus, from eqn (1):

y = 160 - 2(40 yards)

y = 80 yards

and maximum area will be:

Amax = (40 yards)(80 yards)

Amax = 3200 yards squared