Find the global maximum and minimum value of $g(x,y) = x^4 + y^4 -4xy +2$ on the region $D =\{(x,y):0\leq x\leq 3,\ 0\leq y\leq2\}.$
One way to solve this question is first to find the critical points and then consider the boundaries by substituting $x = 0$ or $3$ and $y = 0$ or $2$. Then check the value of max/min and start and endpoints. Among all points, pick the max and min
But can I use the Lagrange multipliers to solve this question? Thanks in advance.
Since $D$ is a rectangle, we'll examine $f$ inside $D$ and on the boundary.
Since $$ f_x(x,y)=4x^3-4y, \quad f_y(x,y)=4y^3-4x $$ solving the system $$ \left\{\begin{array}[ccc] 1f_x(x.y)&=&0\\ f_y(x,y)&=&0 \end{array} \right., $$ we get $$ x=y^3=x^9 $$ i.e. $x=0,\pm1$ and $y=x^3$. So the solutiobs are $(-1,-1), (0,0), (1,1)$. The first pair can be excluded because it isn't inside $D$, the second lies on the boundary of $D$ while the last is an interior point of $D$. In conlusion, $(1,1)$ is the only interior point of $D$ wgich is a critica; poibt of $f$.
Now, let's examine $f$ on the boundary of $D$. It is worth mentioning that the boundary of $D$ consists of four line segments $C_1,C_2, C_3$ and $C_4$ with \begin{eqnarray} C_1 &=&\{(x,0): 0\le x\le 3\}\\ C_2 &=&\{(3,y): 0\le y\le 2\}\\ C_3 &=&\{(x,2): 0\le x\le 3\}\\ C_4 &=&\{(0,y): 0\le y\le 2\} \end{eqnarray}
On $C_1$, $f(x,0)=x^4+2$ is increasing and therefore has am absolute minimum of $f(0,0)=2$ and an absolute maximum of $f(3,0)=83$.
On $C_2$, $f(3,y)=y^4-12y+83$ is decreasing on $0<y<\sqrt[3]{3}$ and increasing $\sqrt[3]{3}< y<2$. Since $f(3,\sqrt[3]{3})=83-9\sqrt[3]{3}<f(3,2)=75<f(3,0)=83$, on $C_2$, $f(3,y)$ has an absolute minimum of $f(3,\sqrt[3]{3})=83-9\sqrt[3]{3}$ and an absolute maximum of $f(3,0)=83$
On $C_3$, $f(x,2)=x^4-8x+18$ is decreasing on $0<x<\sqrt[3]{2}$ and increasing $\sqrt[3]{2}< x<3$. Since $f(\sqrt[3]{2},2)=18-6\sqrt[3]{2}<f(0,2)=18<f(3,2)=75$, on $C_3$, $f(x,2)$ has an absolute minimum of $f(\sqrt[3]{2},2)=18-6\sqrt[3]{2}$and an absolute maximum of $f(3,2)=75$
On $C_4$, $f(0,y)=y^4+2$ is increasing and therefore has am absolute minimum of $f(0,0)=2$ and an absolute maximum of $f(0,2)=18$.
Hence the absolute minimum is $f(1,1)=0$ and the absolute maximum is $f(3,0)=83$.