A Farmer is Enclosing two different breeds of Animals

Question

A farmer needs to enclose two different breads of animals in a rectangular area. The two breads are separated by a fence. One of the lengths are not needed because the field borders a river. Therefore, the farmer will need 1 length section of fence and 3 width sections of fence. See picture below. The farmer has 227 meters of fence and wants the meadow to have as much area as possible. What should the dimensions of the meadow be?

Answer 1

max A=lw subject to l+3w=227

l= 227-3w

maximize w(227-3w) = 227w - 3w^2

Since the title is algebra, I will solve by completing the square.

-3(w^2 - 227w/3 + (227/6)^2) + 3(227/6)^2 =

-3(w - 227/6)^2 + 3(227/6)^2

If w = 227/6, l = 227 - 3(227/6)) = 227/2

A = 227^2/12

Answer: l = 227/2; w = 227/6; A = 227^2/12

(If this were calculus, we would take the derivative of 227w - 3w^2 and set equal to 0, and prove it was a maximum by taking the second derivative and noticing that it was negative everywhere).