Find the absolute extrema of f(x,y)=x^2+2y^2-x bounded by the circular region x^2 + y^2 =

Question

Find the absolute extrema of f(x,y)=x^2+2y^2-x bounded by the circular region x^2 + y^2 =< 4.

jamesj · Answer

Let's call the circular region R,
$$ R = \{ (x,y) \ | \ x^2 + y^2 \leq 4 \} $$

To solve the problem, you do this in two steps.  First, find the extrema of the function in the interior of the region
$$ interior(R) = \{ (x,y) \ | \ x^2 + y^2 < 4 \} $$
You do that by calculating partial derivatives etc.

Then in the second step, examine the function f(x,y) on the boundary of R, $ \ x^2 + y^2 = 4 $.

Which of these steps is giving you trouble?

anonymous · Answer

Both of them.

jamesj · Answer

Ok.  So if you were trying to find the extrema of a function of one variable on the real number line, you would find the derivative and see where it is equal to zero.

What is the generalization of that idea for functions of two variables? You must have covered this in your lectures because otherwise this question is impossible.

jamesj · Answer

?

jamesj · Answer

If you don't know the answer to this, you have to go back to your lecture notes or text book.  This is at the heart of the theory for finding local max and mins of functions in more than one variable.

anonymous · Answer

Well for the inside, you take the critical points and use those to evaluate the function to get extremas.

For the boundaries, you have to express the function in 1 variable using the equations of the sides. Then use that to evaluate the function, then derive it, and check what value of the variable will make the derivative of the function 0. Then evaluate the function using the interval, and the value of the variable you just got.

jamesj · Answer

Ok, so in particular, how do you do find the extrema of f in the interior of R.

anonymous · Answer

For the inside of the circle, what I did was take the gradient of f (in other words, partial derivatives of f with respect to x and y), and equate them to 0. Then I solved for the system of equations. I got ($\frac{1}{2}$,0) here. Using this point to evaluate the function, I get $-\frac{1}{4}$. 

I'm not used to dealing with circular boundaries. So I guess therein lies my problem.

jamesj · Answer

Right, you do know.

For the boundary, 
$$ f(x,y) = x^2 + y^2 - 2x = 4 - 2x $$

anonymous · Answer

Why did you subtract 2x from both sides?

jamesj · Answer

Sorry, I misread the function
$$ f(x,y) = x^2 + 2y^2 - x $$
$$ = (x^2 + y^2) + y^2 - x $$
$$ = 4 + y^2 - x $$

jamesj · Answer

Now as on the boundary
$$ x^2 + y^2 = 4 $$
we also have $ y^2 = 4 - x^2 $, hence
$$ f(x,y) = 4 + (4 - x^2) - x $$

Now find the extrema of that function for $ x \in [-2,2] $

anonymous · Answer

In your former post, how did you proceed from the first line to the second line?

anonymous · Answer

Never mind, I got it.

jamesj · Answer

By the way, this might seem particular to the problem and a bit unsatisfactory.  That's a reasonable response.  The way to get around that is to parameterize the circle in the usual way: with polar coordinates.  Here they would be
$$ x = 2 \cos \theta, \ \ y = 2 \sin \theta $$
Substitute that into the form of f(x,y) and you'll have an equation in theta to analyze. Both methods will give you the same results.  It would be a good exercise to check this. ;-)

anonymous · Answer

Question: Why is it that you can just plug in the equation of the circle into the function? I know that doing so is similar to finding their intersection(s), but I don't see why in this context.

jamesj · Answer

We know on the boundary that x^2 + y^2 = 4.  So why not use that fact to simplify the function?

jamesj · Answer

If you used the polar form of the circle, you would find exactly the same thing, as the cos^2 and sin^2 would simplify.

anonymous · Answer

For the outside, I got the following extrema values: 8.75, 3, -5. Respectively, their x values are -1/2, -2, 2. My problem is how do you take the y values for the corresponding x values, in order to name the point where these extremas occur?

jamesj · Answer

If you're on the boundary, there are two y values for each x value (except x = 2, -2); so use them.

jamesj · Answer

...and those values are the negative of each other.  Hence when they appear in the function as 2y^2, they are equal.

anonymous · Answer

So that means I'm supposed to use y = +/- sqrt(4-x^2), right?

jamesj · Answer

yes

anonymous · Answer

The absolute maximum occurs at $(-\frac{1}{2},\pm\frac{\sqrt{15}}{2})$ and has a value of 8.75, whereas the absolute minimum occurs at (2,0) and has a value of -5.