Question

If $(u,v)$ are the coordinates of the point on the curve $x^3=y(x-4{)}^2$ where the ordinate is minimum, then $uv$ is equal to

Answer 1

The ordinate of any point on the curve is given by $y=\frac{x^3}{(x-4{)}^2}$

$\frac{dy}{dx}=\frac{3x^2}{(x-4{)}^2}-\frac{2x^3}{(x-4{)}^3}=\frac{x^2(x-12)}{(x-4{)}^3}$

$\frac{dy}{dx}=0\Rightarrow x=0\text{or}x=12$

and $\frac{d^{2}y}{d{x}^2}=\frac{(x-4{)}^3(3x^2-24x)-3(x-4{)}^2(x^3-12x^2)}{(x-4{)}^6}=\frac{96x}{(x-4{)}^4}$

Clearly, ${\frac{d^{2}y}{d{x}^2}|}_{x=0}=0$

and ${\frac{d^{2}y}{d{x}^2}|}_{x=12}>0$

Hence $y$ is minimum at $x=12$ and its value is

$y=\frac{(12{)}^3}{(8{)}^2}=27$

Thus $(u,v)=(12,27)$
Hence $uv=324$