Find the maximum and minimum of the function $f(x,y)=xy$ in respect of the curve $x^4+y^4=32$.
Well I defined Lagrange function $L(x,y,\lambda)=xy+\lambda(x^4+y^4-32)$ Then
$L'_x=y+4\lambda x^3=0$
$L'_y=x+4\lambda y^3=0$
$L'_{\lambda}=x^4+y^4-32=0$
However, I have difficulty solving this system. Can you please suggest a way to solve this non-linear system?
On
IMHO, the easiest way is to bring the terms with $\lambda$ on the RHS.
$y+4\lambda x^3=0\Rightarrow y-4\lambda x^3 \qquad (1)$
$x+4\lambda y^3=0\Rightarrow x-4\lambda y^3 \qquad (2)$
Next we exclude the case $x=y=0$ for the next steps.
Dividing (1) by (2) gives
$\frac{y}{x}=\frac{x^3}{y^3}\Rightarrow y^4=x^4$. Inserting the expression for $x^4$ into the third FOC (first order condition).
$$x^4+x^4-32=0\Rightarrow x^4=16$$
This gives 2 real numbers for $x$. I think you can go on by yourself.
We can assume $xy\neq 0 $ and also $\lambda \neq 0$ then from the first two equations we obtain:
and dividing term by term:
$$\frac x y = \frac{x^3}{y^3} \implies x^4=y^4 \implies |x|=|y|$$
and finally from the third equations we obtain:
$$2x^4 = 32 \implies x=\pm 2$$
As an alternative we can plug $y=\sqrt[4]{32-x^4}$ in the original function to obtain:
$$g(x) = f(x,y)= x\sqrt[4]{32-x^4}$$
and study $g'(x)=0$.