Question

The maximum value of $xy$ when $x+2y=8$ is:

• A. 20
• B. 16
• C. 24
• D. 8
• E. 4

Answer 1

Answer: (D)

Given that, $y=\frac{8-x}{2}$
Let $p=xy=x\left(\frac{8-x}{2}\right)$
$=\frac{8x-x^2}{2}$
On differentiating w.r.t. $x,$ we get
$\frac{dp}{dx}=\frac{1}{2}(8-2x)$
Put $\frac{dp}{dx}=0$ for maxima or minima
$\therefore$ $8-2x=0$
$\Rightarrow$ $x=4$
Again $\frac{d^{2}p}{d{x}^2}=\frac{1}{2}(-2)=-1$
${\left(\frac{d^{2}p}{d{x}^2}\right)}_{x=4}=-1<0$
Thus, function is maximum at
$x=4$ and $y=2$
Therefore, maximum value of $p=4\times 2=8$ .