Answer:
400 ft^2
Explanation:
It can be shown that a square area is the most efficient way in which to use fencing. If the area is not square, the area will inevitably be smaller.
Calculus is the tool most often used in higher math to solve optimization problems.
But the same goal can be achieved in this problem by working with constraints:
If x and y are the length and width respectively, then
2x + 2y = 80 ft, or x + y = 40, or x = 40 - y. This is one constraint.
The other constraint involves the area: A = x*y, or A = (40 - y)*y. To maximize this, we need to rewrite (40 - y)*y in standard form:
A = 40y - y^2, or, finally, A = -y^2 + 40 y. The coefficients of this quadratic are -1, 40 and 0; the axis of symmetry is thus
x = -b/ [2a], or, in this case, x = -40/[2*(-1)], or x = 20.
Thus, If x = 20, y = 20 also, proving that the shape of the enclosed yard is that of a square.
Then Mrs. L' 80 feet of fencing is sufficient to construct a 20 ft by 20 ft space, which comes out to a maximum area of 400 ft^2.
40 -