Question

Let $f(x,y)=x{y}^2$ if $x$ and $y$ satisfy $x^2+y^2=9$ then the minimum value of $f(x,y)$ is

• A. $0$
• B. $-3\sqrt{3}$
• C. $-6\sqrt{3}$
• D. $-3\sqrt{6}$

Answer 1

Answer: (C)

$-6\sqrt{3}$

We have,

$f\left(x,y\right)=x{y}^2,x^2+y^2=9y^2=9-x^2\to \left(ii\right)Then,\Rightarrow f\left(x,y\right)=x\left(9-x^2\right)\Rightarrow f\left(x,y\right)=\left(9x-x^3\right)$

$f^{\prime }\left(x,y\right)=9-3x^2=0\Rightarrow 9-3x^2=0\Rightarrow 3\left(3-x^2\right)=0\Rightarrow \therefore x^2=3\Rightarrow x\pm \sqrt{3}and{y}^2=9-3=6$

Then, maximum value of $f\left(x,y\right)=-\sqrt{3}\times 6=-6\sqrt{3}.$

Hence, this is the answer.