$f(x, y)=exp(-3x)+exp(-y)+5x^2y^2$ initial guess $x^0=0, y^0=0$

Question

Qno: Find critical points for $f$.

Can anyone help to understand this and how to solve it?

Do I have to use first derivative test or is there any other numerical method to solve this type of problem ?

$$f_x=-3e^{-3x}+10xy^2=0$$

$$f_y=-e^{-y}+10x^2y=0$$

What will be such $x, y$ to satisfy above equations?

Answer 1

If you are supposed to know Lambert function $$f_y=0\quad \implies \quad y=W\left(\frac{1}{10 x^2}\right)$$ So, it remains to solve for $x$ $$10 \,x\,\Bigg( W\left(\frac{1}{10 x^2}\right)\Bigg)^2-3 e^{-3 x}=0$$

Using graphics, the solutions are close to $0.01$, $0.25$ and $1.67$.

Graphically, the roots are easier to see if you consider the intersections of functions $$g(x)=-3x+\log(3)$$ $$h(x)=\log(10)+\log(x)+2 \log\Bigg( W\left(\frac{1}{10 x^2}\right)\Bigg)$$

Edit

Notice that, if $x$ is "large" $$\Bigg( W\left(\frac{1}{10 x^2}\right)\Bigg)^2=\frac{1}{100 x^4}+O\left(\frac{1}{x^6}\right)$$ and then $$\frac{1}{10 x^3}-3 e^{-3 x}=0 \quad \implies \quad x=-W_{-1}\left(-\frac{1}{\sqrt[3]{30}}\right)\sim 1.61$$

Answer 2

The critical points of $f$ occur where the gradient $\nabla f$ vanishes. This gives a system of two equations in $x,y$.