Finding extrema of multivariable function on a circle

Question

I'm having trouble with finding the extrema of a function on a set which is expressed with a closed circle.

This is the function: $f(x,y)=x^2+y$ , and this is the set: $A=[(x,y)∈R^2: \frac{x^2}4+\frac{y^2}9\le1]$

I know how to find the regular extrema of the function but I'm stuck with the set part of this problem.

I've already calculated that there is no global max/min for $f(x,y)=x^2+y$

Thank you in advance for any help.

Answer 1

HINT

Note that $\frac{x^2}4+\frac{y^2}9\le1$ is not a circle but an ellipse.

To find the extrema:

• find critical points of the function for $\frac{x^2}4+\frac{y^2}9<1$ by $f_x=0$ and $f_y=0$ and determine if necessary whether or not those are max/min/saddle by Hessian matrix
• find extrema points on the boundary $\frac{x^2}4+\frac{y^2}9=1$ by Lagrange multipliers or by substitution of the constraint/boundary into the function equation, that is

$$\frac{x^2}4+\frac{y^2}9=1\implies x^2=4-\frac49y^2$$

$$f(x,y)=x^2+y\implies g(y)=4-\frac49y^2+y$$

for $y\in[-3,3]$.

Answer 2

Since we have $$\frac{\partial f(x,y)}{\partial y}=1\ne 0$$ you must search for extrema on the boundary and you can consider the function $$g(y)=4-\frac{4}{9}y^2+y$$