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?

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

Step-by-step 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