This solution can be confirmed by the substitution method. From 2x + y = 100 we get y = 100−2x, so the problem is reduced to maximizing the unconstrained function h(x) = x(100−2x)=− 2x2 +100x. Since h(x) =− 4x +100 = 0 givesx = 25, and h(x) =− 4 < 0 for all x, this shows that x =25 is a maximum point.