Basically, what this asks you is to maximize the are A=ab where a and b are the sides of the recatangular area (b is the long side opposite to the river, a is the short side that also is the common fence of both corrals). Your maximization is constrained by the length of the fence, so you have to maximize subject to 3a+b=450 (drawing a sketch helps - again, b is the longer side opposite to the river, a are the three smaller parts restricting the corrals) 3a+b = 450 b = 450 - 3a so the maximization max(ab) becomes max(a(450-3a)=max(450a-3a^2) Since this is in one variable, we can just take the derivative and set it equal to zero: 450-6a=0 6a=450 a=75 Plugging back into b=450-3a yields b=450-3*75 b=450-225 b=215 Hope that helps!
