1.      At optimal value of X, ∂Y/∂X = 0

Y = - 2 X ² + 20 X - 20

∂Y/∂X = - 4 X + 20

-4 X + 20 = 0

4 X = 20

X = 20/ 4

   = 5

2.      To find whether the optimal point is a maximum or a minimum point, we find the second derivative. If the second derivative is negative, then it is a maximum point. If the second derivative is positive, then it is a minimum point.

∂Y/∂X = - 4 X + 20

∂ ²Y/∂X ² = - 4

This therefore means that it is a minimum point.