Optimization of ellipse surrounding a circle

[TyroneTheDino]

Homework Statement

Consider the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ that encloses the circle $x^{2}+y^{2}=2x$. Find the values of a and b that minimize the area of the ellipse.

Homework Equations

$Area=ab\pi$

The Attempt at a Solution

I begin by completing the square of the circle equation to get:

$(x-1)^2+y^2=1$

I note that this circle is centered at (1,0). I know that a>b for a minimal area where the ellipse will touch the circle at 2 points, and if that is so, then a=x of the circle.

I know I need to find a quadratic equation for x in terms of a and b by eliminating y^2. Then derive the ellipse and circle equation implicitly, and set them equal to each other.

After an attempt of using these directions i get:

An ellipse equation of
$\frac{x^{2}}{a^2}+\frac{-(x-1)^2+1}{b^2}=1$
and the circle equation: $y^2=-(x-1)^2+1$

I hesitate to derive them because I think I'm missing a concept, but if these equations were correct I would derive them and set them equal to each other to find a relation between a and b.

I am not sure if I am on wrong track, but please don't hesitate to tell me if I am misunderstanding a step.

 

[Ray Vickson]

Homework Statement

Consider the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ that encloses the circle $x^{2}+y^{2}=2x$. Find the values of a and b that minimize the area of the ellipse.

Homework Equations

$Area=ab\pi$

The Attempt at a Solution

I begin by completing the square of the circle equation to get:

$(x-1)^2+y^2=1$

I note that this circle is centered at (1,0). I know that a>b for a minimal area where the ellipse will touch the circle at 2 points, and if that is so, then a=x of the circle.

I know I need to find a quadratic equation for x in terms of a and b by eliminating y^2. Then derive the ellipse and circle equation implicitly, and set them equal to each other.

After an attempt of using these directions i get:

An ellipse equation of
$\frac{x^{2}}{a^2}+\frac{-(x-1)^2+1}{b^2}=1$
and the circle equation: $y^2=-(x-1)^2+1$

I hesitate to derive them because I think I'm missing a concept, but if these equations were correct I would derive them and set them equal to each other to find a relation between a and b.

I am not sure if I am on wrong track, but please don't hesitate to tell me if I am misunderstanding a step.

You have a problem of the form
\begin{array}{l}\min_{a,b} \pi a b\\<br /> \text{subject to}\\<br /> \displaystyle \frac{x^2}{a^2} + \frac{1-(x-1)^2}{b^2} \leq 1 \; \forall x \in [0,2] <br /> \end{array}
This arises because any point $(x,y)$ on the circle must be inside or on the boundary of the elliptical region $x^2/a^2 + y^2/b^2 \leq 1$.

As written, you have infinitely many constraints, one for each $x$ in the range $[0,2]$. In other words, you have infinitely many constraints of the form $f(x,a,b) \leq 1$ for all $x$. Can you think of first solving an auxiliary optimization problem to turn this infinite number of constraints into a single constraint on $a,b$? (Hint: it can be done.)

Then you will have a simpler optimization problem of the form
\begin{array}{l}\min_{a,b} \pi a b \\<br /> \text{subject to} \\<br /> g(a,b) = 0 \end{array}
for some relatively nice function $g(a,b)$. At that point you can apply various methods, such as Lagrange multipliers, to finish the problem.

 

[TyroneTheDino]

You have a problem of the form
\begin{array}{l}\min_{a,b} \pi a b\\<br /> \text{subject to}\\<br /> \displaystyle \frac{x^2}{a^2} + \frac{1-(x-1)^2}{b^2} \leq 1 \; \forall x \in [0,2]<br /> \end{array}
This arises because any point $(x,y)$ on the circle must be inside or on the boundary of the elliptical region $x^2/a^2 + y^2/b^2 \leq 1$.

As written, you have infinitely many constraints, one for each $x$ in the range $[0,2]$. In other words, you have infinitely many constraints of the form $f(x,a,b) \leq 1$ for all $x$. Can you think of first solving an auxiliary optimization problem to turn this infinite number of constraints into a single constraint on $a,b$? (Hint: it can be done.)

Then you will have a simpler optimization problem of the form
\begin{array}{l}\min_{a,b} \pi a b \\<br /> \text{subject to} \\<br /> g(a,b) = 0 \end{array}
for some relatively nice function $g(a,b)$. At that point you can apply various methods, such as Lagrange multipliers, to finish the problem.

How exactly would I solve an auxiliary optimization to turn into a single restraint?

 

[TyroneTheDino]

You have a problem of the form
\begin{array}{l}\min_{a,b} \pi a b\\<br /> \text{subject to}\\<br /> \displaystyle \frac{x^2}{a^2} + \frac{1-(x-1)^2}{b^2} \leq 1 \; \forall x \in [0,2]<br /> \end{array}
This arises because any point $(x,y)$ on the circle must be inside or on the boundary of the elliptical region $x^2/a^2 + y^2/b^2 \leq 1$.

As written, you have infinitely many constraints, one for each $x$ in the range $[0,2]$. In other words, you have infinitely many constraints of the form $f(x,a,b) \leq 1$ for all $x$. Can you think of first solving an auxiliary optimization problem to turn this infinite number of constraints into a single constraint on $a,b$? (Hint: it can be done.)

Then you will have a simpler optimization problem of the form
\begin{array}{l}\min_{a,b} \pi a b \\<br /> \text{subject to} \\<br /> g(a,b) = 0 \end{array}
for some relatively nice function $g(a,b)$. At that point you can apply various methods, such as Lagrange multipliers, to finish the problem.

Also how do we know that the circle and ellipse are in the range [0,2] for x values?

 

[davidmoore63@y]

I think he's trying to say: If the LHS<=1 for all x in [0,2] , then in particular we have Max(over x) of the LHS <=1...

 

[TyroneTheDino]

I think he's trying to say: If the LHS<=1 for all x in [0,2] , then in particular we have Max(over x) of the LHS <=1...

I don't think I'm following that relationship either. How am I supposed to know the relationship between the ellipse and the circle.

 

[TyroneTheDino]

I think he's trying to say: If the LHS<=1 for all x in [0,2] , then in particular we have Max(over x) of the LHS <=1...

You have a problem of the form
\begin{array}{l}\min_{a,b} \pi a b\\<br /> \text{subject to}\\<br /> \displaystyle \frac{x^2}{a^2} + \frac{1-(x-1)^2}{b^2} \leq 1 \; \forall x \in [0,2]<br /> \end{array}
This arises because any point $(x,y)$ on the circle must be inside or on the boundary of the elliptical region $x^2/a^2 + y^2/b^2 \leq 1$.

As written, you have infinitely many constraints, one for each $x$ in the range $[0,2]$. In other words, you have infinitely many constraints of the form $f(x,a,b) \leq 1$ for all $x$. Can you think of first solving an auxiliary optimization problem to turn this infinite number of constraints into a single constraint on $a,b$? (Hint: it can be done.)

Then you will have a simpler optimization problem of the form
\begin{array}{l}\min_{a,b} \pi a b \\<br /> \text{subject to} \\<br /> g(a,b) = 0 \end{array}
for some relatively nice function $g(a,b)$. At that point you can apply various methods, such as Lagrange multipliers, to finish the problem.

I know this problem is emulative of https://www.physicsforums.com/threa...area-of-an-ellipse-enclosing-a-circle.270437/ this one however I am just getting into multivariable differentiation so this is very confusing to me.

 

[TyroneTheDino]

How about since
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$(x-1)^2+y^2=1$

Can I set each side equal to each other or should I solve for y^2 of the circle equation to plug into the ellipse equation.

 

[Ray Vickson]

Also how do we know that the circle and ellipse are in the range [0,2] for x values?

Draw a circle of radius 1, centered at (1,0). What are the x-values you can have on the circle?

I never, anywhere, said that the ellipse can extend only over 0 <= x <= 2; in fact, it can extend out to x = 10 million if you want it to. All I said was that the part of the ellipse lying between x = 0 and x = 2 must be outside the circle, and I put that mathematically in terms of an inequality constraint.

As for telling you what to do next: I am not allowed to do that, by PF rules.