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6. Application of derivatives — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths6. Application of derivatives1 Mark
MCQ
The minimum value of $\left( {{x^2} + {{250} \over x}} \right)$ is
  • ✓
    $75$
  • B
    $50$
  • C
    $25$
  • D
    $55$
✓

Answer

Correct option: A.
$75$
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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