Find the maximum and minimum values of x 3 + 3xy 2 – 3x 2 – 3y 2 + 7.

Let, $f(x,y)=x^3+3xy^2-3x^2-3y^2+7 \\ \; \\ $

The stationary points are given by $\dfrac{\partial u}{\partial x}=0$ and $\dfrac{\partial u}{\partial y}=0$

$\\ \; \\ \therefore \dfrac{\partial u}{\partial x} \;=\; 3x^2+3y^2-6x=0 \; \; \; \therefore x^2+y^2-2x=0 \; \; \ldots(i) \\ $

Also, $ \\ \; \\ \therefore \dfrac{\partial u}{\partial y} …