Optimization - minimize area of an ellipse enclosing a circle

In summary, the book describes a problem in which an ellipse must enclose the circle x2+y2=2y. If a>b, then b=y of the circle. However, the author discovered that if the ellipse is already tangent to the circle in two places, then b cannot be decreased any further to create the smallest possible area. The solution where the ellipse hits the circle at only at (0,2) isn't optimal. The author suggests finding a way for the ellipse to hit the circle tangentially at two points of equal y value, and then minimizing with respect to y.
  • #1
ColdSanctuary
5
0
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.
 
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  • #2
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
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
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
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
Welcome to PF!

ColdSanctuary said:
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. …

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. :wink:
 
  • #7
ColdSanctuary said:
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.

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
Oh wow!Thank you! I tried the method you told me (Dick) and it worked! Thank you so much!
 
  • #9
ColdSanctuary said:
Oh wow!Thank you! I tried the method you told me (Dick) and it worked! Thank you so much!

Good job!
 

FAQ: Optimization - minimize area of an ellipse enclosing a circle

1. How do you find the optimal area of an ellipse enclosing a circle?

The optimal area of an ellipse enclosing a circle can be found by using the formula A = πr2, where r is the radius of the circle. This formula represents the area of a circle, which is the smallest possible area that can enclose the circle.

2. What is the relationship between the area of an ellipse and the area of a circle?

The area of an ellipse is always greater than or equal to the area of a circle. This is because an ellipse is a stretched out version of a circle, and the distance between the two foci of an ellipse is always greater than the radius of a circle.

3. How does changing the eccentricity of an ellipse affect its area?

The eccentricity of an ellipse is a measure of how "stretched out" the ellipse is. As the eccentricity increases, the ellipse becomes more elongated and its area increases. This means that for a given circle, an ellipse with a higher eccentricity will have a larger area.

4. Can the area of an ellipse enclosing a circle be minimized?

Yes, the area of an ellipse enclosing a circle can be minimized. This can be achieved by choosing an ellipse with a semi-major axis that is equal to the radius of the circle and a semi-minor axis that is equal to half the radius of the circle. This results in an ellipse with an eccentricity of 1, which is a circle with the smallest possible area that can enclose the given circle.

5. What are some real-world applications of minimizing the area of an ellipse enclosing a circle?

One potential application of minimizing the area of an ellipse enclosing a circle is in the design of satellite orbits. By minimizing the area of the ellipse that represents the orbit, less fuel is required to maintain the orbit, resulting in cost savings for space missions. Another application could be in the design of bridges, where minimizing the area of the ellipse enclosing the support structure can lead to more efficient and cost-effective designs.

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