# Optimization - minimize area of an ellipse enclosing a circle

1. Nov 8, 2008

### ColdSanctuary

This is how the book describes the problem:
If the ellipse x2/a2+y2/b2=1 is to enclose the circle x2+y2=2y, what values of a and b minimize the are of the ellipse?

First of all I completed the square for the second equation and I got: x2+(y-1)2=1. I isolated the x2 and substituted it into the ellipse formula because after drawing some diagrams, I realized that if b>a, for a minimal area the ellipse will touch the circle at two points. If a>b, then b=y of the circle.

I'm really lost though... i don't know what to do from here. I don't even know where to start.
I tried solving (y-1)2/a2+y2/b2=1 but it got messy.

2. Nov 8, 2008

### Dick

If you've drawn some diagrams, you probably figured out that the circle is centered at (0,1) and has radius 1. The ellipse is centered at (0,0). So an ellipse of minimal area is going to pass through (0,2), right? Otherwise you could shrink b a little and make the area smaller. What does this tell you about the value of b?

3. Nov 8, 2008

### ColdSanctuary

Yeah, that means that b=2. However, I also have the answers at the back of the book and it says that a=sqrt(6)/2 and b=3sqrt(2)/2.
I don't know exactly how to combine the formulas and use those to create a system of equations... I tried integrating 1/4 of the area of the ellipse with respect to y, and I got the 1/4 area function with y, a and b in it. But if I derive it again and find the critical points, won't I go back to the original function? That is, A/4?

4. Nov 8, 2008

### Dick

Gack! The answer is right. I'm wrong. Sorry! You can't shrink b if the ellipse is already tangent to the circle in two places. The solution where the ellipse hits the circle at only at (0,2) isn't optimal. Looks like you have to figure out how the ellipse can hit the circle tangentially at two points of equal y value, and then minimize with respect to y. Or with respect to an angle. Can I have another look at this tomorrow? It's harder than I thought, and I didn't think it was all that easy to begin with.

5. Nov 9, 2008

### ColdSanctuary

Sure, no problem. If it helps, I'm doing this in a Multivariable calculus course, right after partial derivatives section. I'll try again and I'll also ask my classmates. Thank you, I'll check this page again later.

6. Nov 9, 2008

### tiny-tim

Welcome to PF!

Hi ColdSanctuary! Welcome to PF! :smile

Hint: change the origin to (0,1), then you're looking for the minimum distance from the origin, and you want that to be 1.

7. Nov 9, 2008

### Dick

The missing ingredient is that not only are the y values equal where the circle and ellipse touch, dy/dx is also the same, since they will be tangent. So keep going the way you started. Find a quadratic equation for y in terms of a and b by eliminating x^2. Then implicitly differentiate the equations for the circle and the ellipse. That will give you another equation for y in terms of a and b. If you set those two equal, you will find a pretty simple relation between a and b. Use that to turn the area equation pi*a*b into a function of a single variable and minimize. I've tried it and I endorse this method. You just have to plow through it. It's not as bad as it sounds.

8. Nov 9, 2008

### ColdSanctuary

Oh wow!!Thank you! I tried the method you told me (Dick) and it worked! Thank you so much!

9. Nov 9, 2008

Good job!