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

Options

A. `x = (1)/(5)` ` Y = 4`
B. `x = (1)/(5)` ` y = 2`
C. `x = (2)/(5)` ` Y = 4`
D. `x = (1)/(5)` ` Y = 14`

Answer

A

Explanation

(a) 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) y)/(dx^(2))` has positive 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`.
(b) `y=9- (x-3)^(2) = 9-x^(2) - 9 + 6x`
or `y=6x-x^(2)`
`:. (dy)/(dx) = 6-2x`
For minimum or maximum value of `y` we will substitute `(dy)/(dx) = 0`.
or `6-2x=0` or `x=3`
To check whether value of `y` is maximum or minimum at `x=3` we will have to check whether `(d^(2) y)/(dx^(2))` is positive or negative. `(d^(2) y)/(dx^(2)) = -2`
or `(d^(2) y)/(dx^(2))` is negative at `x =3`. Hence, value of `y` is maximum. This maximum value of `y` is,
`y_(max) = 9-(3-3)^(2) = 9`