MCQ
The maximum value of $xy $ when $x + 2y = 8$ is
- A$20$
- B$16$
- C$24$
- ✓$8$
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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