The area of the garden enclosed by the fencing is
A(x, y) = xy
and is constrained by its perimeter,
P = x + 2y = 200
Solve for x in the constraint equation:
x = 200 - 2y
Substitute this into the area function to get a function of one variable:
A(200 - 2y, y) = A(y) = 200y - 2y²
Differentiate A with respect to y :
dA/dy = 200 - 4y
Find the critical points of A :
200 - 4y = 0 ⇒ 4y = 200 ⇒ y = 50
Compute the second derivative of A:
d²A/dy² = -4 < 0
Since the second derivative is always negative, the critical point is a local maximum.
If y = 50, then x = 200 - 2•50 = 100. So the farmer can maximize the garden area by building a (100 ft) × (50 ft) fence.
