Answer:
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)
<u>Amax = 3200 yards squared</u>
