# Finding the max and min values for a surface on region R

Surface $$\ f(x,y)=x^2+4y^2-x+2y$$ on the region bounded by $$\ x^2+4y^2=1$$

Finding the critical points of the surface within the region was easy enough, I found a minimum at $\ (\frac{1}{2},\frac{-1}{4})$. What I'm having trouble with is finding the critical points along the boundary. I can plug in the equation for the region into the surface equation, but that leaves me with a square root.

## 2 Answers

Hint: $f(x,y) = 1-x+2y$, and using Cauchy-Schwarz inequality: $(-x+2y)^2 = (1\cdot (-x)+1\cdot (2y))^2\leq (1^2+1^2)(x^2+4y^2)=2$. Can you take it from here ?

• We haven't covered the Cauchy-Schwarz inequality. Is that the only means of solving this problem? – user296558 Dec 13 '15 at 1:45

$x^2+4y^2=1$ is an ellipse. If $g(t)$ is the parametrization of the ellipse, then $f(g(t))$ are the values of $f$ along the ellipse (the boundary). At that point you are left with a single variable optimization problem.