Question

Find the absolute extrema of f(x, y) = x^(2).y − 2.x + y on the
closed triangle T with vertices (1, 0),(5, 0) and (1, 4).

Answer 1

$\frac {\partial f(x,y)} {\partial x}=2xy-2$

$\frac {\partial f(x,y)} {\partial y}=x^2+1$

We should solve the system $\begin {cases} \frac {\partial f(x,y)} {\partial x}=0 \\ \frac {\partial f(x,y)} {\partial y}=0 \end {cases}$ $\iff$ $\begin {cases} 2xy -2=0 \\ x^2+1=0 \end {cases}$

This system has no solution. It means that there is no extremum point at all.

Our triangle T is bounded by these lines: $(x=1; y=-x+5; y=0)$

We must check how the surface behaves on these lines (triangle borders).

1. If $y=0$ then $f=-2x$ is a straight line. It has no extremum point.
2. If $x=1$ then $f=2y-2$ is a straight line too.
3. If $y=-x+5$ then $f=-x^3+5x^2-3x+5$

$f'=-3x^2+10x-3$

$f'=0$ when $x=1/3$ or $x=3$

$x=1/3$ does not lie in the region bounded by the triangle.

if $x=3$ then $y=2$

Now we have the following picture:

M1, M2, M3 are triangles vertices and the point M4 with coordinates (3,2) .

Let's find values of $f(x,y)$ at these points.

M1: $f(1, 0) = -2$

M2: $f(5, 0) = -10$

M3: $f(1, 4)=6$

M4: $f(3, 2)= 14$

Finally, $\max \underset {T}{f(x,y)}=f(3,2)=14,$ $\min \underset {T}{f(x,y)}=f(5,0)=-10$