Question

I need to finf the Global Max/Min points of the function f(x,y)=(x^2)*y + y^2 - 2*y +1 (D: -6<=x<=6 & -6<=y<=6) I have already found the Interior critical Points but I am getting stuck on how to check the boundary points. Thanks for any help :)

Answer 1

Those are the four boundary points. You need to plug each into f(x,y) to find max/min along with critical points inside the boundary. You may have to test the 4 points where it crosses the x/y axes too.

Answer 2

But do I have to do like the derivative and let it equal to zero?

Answer 3

That's what I have seen a lot of examples do:
at x=-6
\[f(-6,y)=36y + y ^{2}-2y+ 1\]
Find Crit Pts of y: (derivative with respect to y = 0
\[2y + 34 = 0\]
then solve for y:
\[y=-17\]
So one critical point will be (-6,-17) but isn't this outside the domain?

Answer 4

like question 2 on this pdf

Answer 5

Oh, I thought you just wanted to CHECK the boundary points for min/max.

If you want to find the critical points along the boundary, then:
\[\large
f(x,y)=(x^2)y + y^2 - 2y +1 \\ \large
\frac{\partial f}{\partial x} = 2xy \\ \large
\frac{\partial f}{\partial y} = x^2 + 2y - 2
\]Boundary lines are: \(\large x = -6;~~~x=6;~~~y=-6;~~~y=6\)
\[\large
\left[\frac{\partial f}{\partial y}\right]_{x=-6} = (-6)^2 + 2y - 2 = 2y+34=0;
~~~y = -17. \\ \text{Outside the domain. Ignore.} \\ \text{ } \\ \large

\left[\frac{\partial f}{\partial y}\right]_{x=+6} = (6)^2 + 2y - 2 = 2y+34=0;
~~~y = -17. \\ \text{Outside the domain. Ignore.} \\ \text{ } \\ \large

\left[\frac{\partial f}{\partial x}\right]_{y=-6} =2x(-6) = -12x = 0;~~~x = 0.\\ \large
(0,-6) \normalsize ~~~\text{Critical point on the boundary.} \\ \text{ } \\ \large

\left[\frac{\partial f}{\partial x}\right]_{y=6} =2x(6) = 12x = 0;~~~x = 0.\\ \large
(0,6) \normalsize ~~~\text{Critical point on the boundary.} \\
\]Evaluate f(x,y) at internal critical points, critical points on the boundary, and the four corners: (6,6); (-6,6); (-6,-6), (6,-6) to determine the global min/max points and its values.