Calc4 Max/Min Problem

[shizzy]

f(x,y) = x^2 + 2y^2 - x
bounded by R where R is the circular region x^2 + y^2 <= 4

Find the absolute maximum and minimum.

Well, I know you need to do three steps here:
a. Find interior critical points
b. Find exterior(boundary) critical points
c. Check all critical points by plugging into f(x,y) to determine min/max values. I think you also check the boundary itself but if you do I don't know how to do that here.

Here I will use notation Fx = partial derivative with respect to x
Fy = partial derivative with respect to y

so :

a. Fx = 2x - 1 and Fy = 4y
solving for the critical points by setting these equal to zero and solving simultenously yields the crit. point (1/2, 0)

b. This is where I don't know what to do. On a similar problem using a triangle as the boundary you found the line segments of the triangle and used the y=mx +b to check for critical points. Then if not you simply added the boundary points to the list of points to check. But here the region is circular so...am I suppose to use the "corners" of the circle? Or what? I'll try that and see what happens in a bit....

c. I get the minimum at -1/4 but not the maximum since it must be based on the boundary area which I'm not sure how to do.

Any help would be awesome. Thanks!!

 

[arthur ohlsten]

this might help
x^2+y^2 =4 is a circle of radius 2 centered at 0,0

x^2-x+2y^2=0 complete the square
x^2-x+1/4 +2y^2 -1/4 =0
[x-1/2]^2 +2y^=1/4 divide both sides by1/44
[x-1/2]^2/ 1/4 +y^2/ 1/8 =1 a ellipse center at 1/2.0
major axis from 0 to 1 along x axis
minor axis + sqrt2/4 -sqrt2/4

max at x=1/2 y= sqrt2 /4
min at x=1/2 y=-sqrt4/2

please check for errors
Arthur

 

[shizzy]

sorry about that, it turns out my teacher assigned more problems than he meant to. So he went over it today and it makes sense now, for anyone who wants to know how to solve this for some reason, here is how:

Basically min/max problems are solved by setting the first derivative equal to zero. On bounded problems you need to check the interior of the function as well as the boundary. So..

Step 1-- CHeck the interior:

Fx = 2x -1 ==> x = 1/2
Fy = 4y ==> y = 0

So this is our first critical point (1/2,0). Evaluating this in the original function yields -1/4. This will end up being the minimum value.

Step 2-- Check the boundary:

I think what we are doing here is getting the function that we're interested in, in terms of the boundary. So we do some substitution.

R==> x^2 + y^2 <= 4
..make it an equality
x^2 + y^2 = 4

I just multiply everything by 2 to get 2y^2 = 8 - 2x^2

Then you can plug this into the original to get:

g(x) = -x^2 -x + 8

Now we do a simple calc1 min/max on this:

g'(x) = -2x - 1 ==> x = -1/2
so here we have x = -1/2,-2,2 since we hafta check the boundary values as well(hence -2 and +2). We just plug these values into g(x) which yields: 33/4, 6, 2 respectively. So all of our critical points/boundary points yielded:

-1/4, 33/4, 6, 2 so clearly the min is -1/4 and the max is 33/4(which is the answer).