7.4-3

Show that the expression q^2 + (n -q -1)^2 achieves a maximum over q = 0. 1... n - 1 when q = 0 or q = n 1. $$F(n) = q^2 + (n -q -1)^2$$ if we take derivative $$F'(n) = 2q -2 (n -q -1)$$ $$ 2q -2 (n -q -1) = 0$$ $$ 2q = 2 (n -q -1) $$ $$ 2q = (n -1) $$ $$ q = (n -1)/2 $$ the above value can be max or min depending on second derivate.Take second derivative F'(n) $$F''(n) = 2 +2 $$ $$F''(n) = 4 $$ second derivative is positive it means this is a minimum value. If we graph above equation it will be a downward-convex. The maximum for a downward convex occurs at its endpoints and minimum at its middle.