find maximum and minimum values of f on R bounded by ellipse

Question

I am having trouble with a problem in my textbook that is asking for the maximum and minimum values of

$f(x,y) = x^2 +4y^2 -x + 2y$ bounded by the ellipse $x^2 +4y^2 = 1$

I know how to find the extreme and saddle points of a function, but am unsure of how to proceed when a boundry is added.

I started by setting $fx$ and $fy$ of $f(x,y) = 0$ and found the point $P(1/2,-1/4)$ using the discriminant. I found that this point is a minimum equal to

$$f(1/2,-1/4) = -1/2$$

now I am not sure how the boundaries factor in. I graphed both functions in a graphing tool and $f(x,y)$ is an elliptical parabaloid, and the ellipse is going vertically through the parabaloid

now im assuming that the first point that I found is the very bottom of $f(x,y)$, but there are places where the cylinder crosses the function. the intersection near the bottom cant be lower than the minimum point P of the function, so I know I am only looking for a max point where the two functions intersect.

aside from finding out how to find the max point, I would appreciate a way for me to find out how many critical points I am looking for when I dont have a graphing calculator available during a test.

Answer 1

Use Lagrange Multipliers, the theorem says that if $P$ is a maximum or minimum of $f$ restricted to the level surface $g$ then there exists $\lambda \in \mathbb{R}$ such that $\nabla f_P=\lambda \nabla g_P$.

Applying this on your problem you get $2x-1=2x\lambda$ and $4y+1=4y\lambda$.

And playing a bit with these you get $x=-2y$. And then you can use that $(x,y)$ is a point of the ellipse. And there you got your candidates.

To check whether they are minimum or maximum just evaluate of $f$

Answer 2

With

$$ f(x,y) = x^2+4y^2-x+2y,\ \ g(x,y) = x^2+4y^2-1 = 0 $$

Taking the lagrangian

$$ L = f(x,y) + \lambda g(x,y) $$

the stationary points are the solutions for

$$ \nabla L = \left\{\begin{array}{rcl} 2x+2\lambda x - 1 & = & 0\\ 8y+8\lambda y + 2 & = & 0\\ x^2+4y^2-1& = & 0 \end{array}\right. $$

The solutions are $x=-\frac{\sqrt2}{2}, y=\frac{\sqrt2}{4}, \lambda = -\frac{1}{2}(2+\sqrt2)$ and $x=\frac{\sqrt2}{2}, y=-\frac{\sqrt2}{4}, \lambda = -\frac{1}{2}(\sqrt2-2)$

The qualification is done according to $f\circ g = 1-x\pm \sqrt{1-x^2}$

The maximum is at $x=-\frac{\sqrt2}{2}, y=\frac{\sqrt2}{4}$ and the minimum at $x=\frac{\sqrt2}{2}, y=-\frac{\sqrt2}{4}$

Attached a plot with the surfaces, and also the qualification