Question

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

Answer 1

(a) Minimum value is 4.

(b) Maximum value is 9.

(a) y = 25x² + 5 - 10x

We differentiate the equation with respect to x.

• $\frac{dy}{dx} = 50x - 10$

To find critical point, we take  $\frac{dy}{dx} = 0$

• 50x - 10 = 0
• x = 1/5

To find maxima or minima, we again differentiate w.r.t. x

• $\frac{d^{2} y}{dx^{2} } = 50$ which is positive for x = 1/5
• This means y has minima at x = 1/5

To find minimum value we substitute x=1/5 in the given equation.

• $y = 25.\frac{1}{5^{2} } +5 -10.\frac{1}{5}$
• y = 1 + 5 - 2
• y = 4 is the minimum value.

(b) y = 9-(x-3)²

• y = 9 - [ x² - 6x + 9]

• y = 6x - x²

Differentiating w.r.t. x,

• $\frac{dy}{dx} = 6 - 2x$

To obtain critical point ,

• 6 - 2 x = 0
• x = 3

To find whether maxima or minima we again differentiate w.r.t. x

• $\frac{d^{2}y }{dx^{2} } = 0-2$ = -2 which is negative for x=3
• y has maximum value at x=3

Substituting x = 3 in given equation

• y = 9 -(3-3)²
• y = 9 is the maximum value

Answer 2

Given :

a) $y = 25x^{2} + 5 - 10 x$

b) $y = 9 - (x - 3)^{2}$

To find :

Maxima and minima values of the given functions.

Solution :

(a) For maximum and minimum value, we can put    $\frac{dy}{dx} = 0$

or  $\frac{dy}{dx} = 50x - 10 = 0$

∴$x = \frac{1}{5}$

Further, $\frac{d^{2}y }{dx^{2} } = 50$

or $\frac{d^{2}y }{dx^{2} }$  has positive value at , Therefore, y has minimum value at $x = \frac{1}{5}$

Substituting in given equation, we get

y min= $25( \frac{1}{5}) ^{2} + 5 -10 ( \frac{1}{5} ) = 4$

(b) $y = 9 -( x-3)^{2} = 9 - x^{2} - 9 + 6x$

    or

    $y = 6x - x^{2}$

  ∴$\frac{dy}{dx} = 6 - 2x$

For minimum or maximum value of y we will substitute $\frac{dy}{dx} = 0$

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  $\frac{d^{2}y }{dx^{2} }$   is positive or negative. $\frac{d^{2}y }{dx^{2} } = - 2$

or ,

 is negative at x=3. Hence, value of y is maximum. This maximum value of y is,

y max= $9 - ( 3 - 3)^{2} = 9$

Hence ,  the maxima and minima values of the given fuctions is 9 and 3.