Find maximum or minimum values of the functions
(a) `y =25 x^(2) + 5 - 10 x`
(b) ` y = 9 - (x - 3)^(2)`

Explanation

For maximum and minimum value, we can put
`(dy)/(dx) = 0 " or "(dy)/(dx) = 50x - 10 = 0`
`:. X = 1/5 " Further," (d^2 y)/(dx^2) = 50`
or `(d^2)/(dx^2)` has positive value at `x = 1/5`
Therefore, y has minimum value at `x = 1/5`. Therefore, `y` has minimum value at `x =
1/5` .
Substituting `x = 1/5` in given equation, we get
`y_("min") = 25(1/5)^(2) + 5 -10(1/5) = 4`.