Question

if xy(y-x) = 16 then y has a minimum value when x=

• A. 1
• B. 3
• C. 2
• D. 4

Answer 1

The correct option is D 4

$xy(y-x)=16$
$x{y}^2-x^{2}y=16$
$y^2-xy-\frac{16}{x}=0$
$(y-\frac{x}{2}{)}^2-\frac{x^2}{4}-\frac{16}{x}=0$
$y=\frac{x}{2}\pm \sqrt{\frac{x^2}{4}+\frac{16}{x}}$
$y^{\prime }=\frac{1}{2}\pm \frac{1}{2}\left(\frac{\frac{2x}{4}-\frac{16}{x^2}}{\sqrt{\frac{x^2}{4}+\frac{16}{x}}}\right)$
$-1=\pm \left(\frac{\frac{x}{2}-\frac{16}{x^2}}{\sqrt{\frac{x^2}{4}+\frac{16}{x}}}\right)$
$\frac{16}{x}=\frac{256}{x^4}-\frac{16}{x}$
$x^3=8$
$x=2$
& $y=4$