Answer
The maximum values are at points $(-2,1)$ and $(2,1)$ and the minimum values are at points $(2,-1)$ and $(-2,-1)$.
Work Step by Step
We need to find the maximum and minimum values of function $f(x, y)=x^2y$ given the constraint $x^2+y^4=5$.
$f_x=\lambda g_x$
$f_y=\lambda g_y$
$2xy=\lambda 2x$
$x^2=\lambda 4y^3$
$x^2+y^4=5$.
From the first equation, $y=\lambda$. Plugging into second equation, we get $x^2=4y^4$. From the third equation, we get $4y^4+y^4=5$, which means $y=-1$ or $y=1$ and $x=\pm2$.
Therefore, we have the four points, $(-2,-1)$, $(2,-1)$, $(-2,1)$, and $(2,1)$.
$f(-2,-1)=(-2)^2(-1)=-4$
$f(2,-1)=(2)^2(-1)=-4$
$f(-2,1)=(-2)^2(1)=4$
$f(2,1)=(2)^2(1)=4$
Thus, the maximum values are at points $(-2,1)$ and $(2,1)$ and the minimum values are at points $(2,-1)$ and $(-2,-1)$.
