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STD 12 - 6. Application of derivatives — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 6. Application of derivatives1 Mark
MCQ
The maximum value of  $xy $ when $x + 2y = 8$ is
  • A
    $20$
  • B
    $16$
  • C
    $24$
  • ✓
    $8$
✓

Answer

Correct option: D.
$8$
d
(d) $x + 2y = 8$, $y = \frac{{8 - x}}{2}$

Now $f(x) = xy = x.\frac{{(8 - x)}}{2} = 4x - \frac{{{x^2}}}{2}$

$\therefore $ $f'(x) = 4 - x$

For extremum,$f'(x) = 0$

$\therefore $ $x = 4$ and $ y = 2.$

Also $f''(x) = - 1 < 0$

So, maximum value of $xy = 4 \times 2 = 8$.

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