First slide

Application of diffrentiation

Question

Find maximum or minimum values of the functions
$y = 25 x^{2} + 5 - 10 x$

Moderate

A

2
B

4
C

6
D

8

Solution

For maximum and minimum value, we can put $\frac{d y}{d x} = 0$
$\frac{d y}{d x} = 50 x - 10 = 0$
$x = \frac{1}{5}$
Further,
$\frac{d^{2} y}{d x^{2}} = 50$
or $\frac{d^{2} y}{d x^{2}}$ has positive value at $x = \frac{1}{5}$ , Therefore, y has minimum value at $x = \frac{1}{5}$ .
Substituting $x = \frac{1}{5}$ in given equation, we get
$y_{\min} = 25 {(\frac{1}{5})}^{2} + 5 - 10 (\frac{1}{5}) = 4$

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