The minimum value of ( x^2 + 250 x ) is - Vidyadip

Question

The minimum value of $\left( {{x^2} + {{250} \over x}} \right)$ is

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Answer

a
(a) Let $y = f(x) = \left( {{x^2} + \frac{{250}}{x}} \right)$,

$\therefore$ $\frac{{dy}}{{dx}} = f'(x) = 2x - \frac{{250}}{{{x^2}}}$

Put $f'(x) = 0$==>$2{x^3} - 250 = 0$==>${x^3} = 125$==> $x = 5$

Again,$\frac{{{d^2}y}}{{d{x^2}}} = f''(x) = 2 + \frac{{500}}{{{x^3}}}$.

Now $f''(5) = 2 + \frac{{500}}{{125}} > 0$

Hence at $x = 5$. The function will be minimum.

Minimum value $f(5) = 25 + 50 = 75$.

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