1. ## Optimisation question

Hello Forumites,
A sphere of radius a is inscribed in a right circular cone, with the sphere touching the base of the cone. Author wants me to find the radius and height of the cone if its volume is minimum.
Answer:-Sphere's volume=$\frac43*\pi*a^3$. It's first derivative is w.r.t.r $4*\pi*a^2$. How sould i use here co-ordinates geometry or lagrange multipliers?

2. ## Re: Optimisation question

Originally Posted by Vinod
Hello Forumites,
A sphere of radius a is inscribed in a right circular cone, with the sphere touching the base of the cone. Author wants me to find the radius and height of the cone if its volume is minimum.
Answer:-Sphere's volume=$\frac43*\pi*a^3$. It's first derivative is w.r.t.r $4*\pi*a^2$. How sould i use here co-ordinates geometry or lagrange multipliers?
I assume that it is the volume of the cone that is to be a minimum. You don't need the volume of the sphere. I would let $h$ and $r$ be the unknown height and radius of the cone. Then I would use the geometry to get a relation between $h$ and $r$. Then in the equation $V = \frac 1 3 \pi r^2h$ of the cone substitute $r$ in terms of $h$ and solve it as a one variable optimization problem.

3. ## Re: Optimisation question

Originally Posted by Walagaster
I assume that it is the volume of the cone that is to be a minimum. You don't need the volume of the sphere. I would let $h$ and $r$ be the unknown height and radius of the cone. Then I would use the geometry to get a relation between $h$ and $r$. Then in the equation $V = \frac 1 3 \pi r^2h$ of the cone substitute $r$ in terms of $h$ and solve it as a one variable optimization problem.
You need the sphere to fit inside the cone, so you definitely need the radius of the sphere.

4. ## Re: Optimisation question

Originally Posted by Walagaster
I assume that it is the volume of the cone that is to be a minimum. You don't need the volume of the sphere. I would let $h$ and $r$ be the unknown height and radius of the cone. Then I would use the geometry to get a relation between $h$ and $r$. Then in the equation $V = \frac 1 3 \pi r^2h$ of the cone substitute $r$ in terms of $h$ and solve it as a one variable optimization problem.
Hello,
How do you use the geometry to get a relation between h andr? $h=\frac{3V}{\pi*r^2}$

5. ## Re: Optimisation question

Originally Posted by SlipEternal
You need the sphere to fit inside the cone, so you definitely need the radius of the sphere.
Hello,
Radius of sphere is a. So h of cone consist of 2a+say z. How can we utilise this information to answer this question?

6. ## Re: Optimisation question

Take a cross section of the cone and sphere (through the center of the sphere). It will look like a circle inscribed in a triangle. We want to figure out the $z$ you mention. Drop a line segment from the top of the cone to the base of the cone that goes through the center of the sphere (this will be an angle bisector of the triangle that goes through the center of the circle and bisects the base of the triangle by symmetry). Now, you have two right triangles. You may want to draw it. The circle will touch the hypotenuse of this right triangle at a right angle. So, if you draw the radius to the point where the circle touches the triangle, you wind up with a similar triangle to the original right triangle. Let's label these points. At the top of the triangle, you have point A. At the base of the triangle, you have points B on the left and C and the right. The angle bisector we drew in bisects BC at point D. The center of the circle is point E. The circle touches the right side of the triangle at point F. Once you have all of this drawn, it should be apparent that:

$$\dfrac{EF}{AF} = \dfrac{DC}{AD}$$

Plugging in what you know:

$$\dfrac{a}{AF} = \dfrac{r}{2a+z}$$

Solving for $AF$ gives:

$$AF = \dfrac{a(2a+z)}{r}$$

Now, by the pythagorean theorem, we have:

$$(a+z)^2 = a^2+\left(\dfrac{a(2a+z)}{r}\right)^2$$

Solve for $z$.

7. ## Re: Optimisation question

Originally Posted by SlipEternal
Take a cross section of the cone and sphere (through the center of the sphere). It will look like a circle inscribed in a triangle. We want to figure out the $z$ you mention. Drop a line segment from the top of the cone to the base of the cone that goes through the center of the sphere (this will be an angle bisector of the triangle that goes through the center of the circle and bisects the base of the triangle by symmetry). Now, you have two right triangles. You may want to draw it. The circle will touch the hypotenuse of this right triangle at a right angle. So, if you draw the radius to the point where the circle touches the triangle, you wind up with a similar triangle to the original right triangle. Let's label these points. At the top of the triangle, you have point A. At the base of the triangle, you have points B on the left and C and the right. The angle bisector we drew in bisects BC at point D. The center of the circle is point E. The circle touches the right side of the triangle at point F. Once you have all of this drawn, it should be apparent that:

$$\dfrac{EF}{AF} = \dfrac{DC}{AD}$$

Plugging in what you know:

$$\dfrac{a}{AF} = \dfrac{r}{2a+z}$$

Solving for $AF$ gives:

$$AF = \dfrac{a(2a+z)}{r}$$

Now, by the pythagorean theorem, we have:

$$(a+z)^2 = a^2+\left(\dfrac{a(2a+z)}{r}\right)^2$$

Solve for $z$.
Hello, I got $z=\frac{a}{r}\sqrt{r^2+(2a+z)^2}-a$. Now, what to do? Because equation is not in one variable form to differentiate.

8. ## Re: Optimisation question

Originally Posted by SlipEternal
$$(a+z)^2 = a^2+\left(\dfrac{a(2a+z)}{r}\right)^2$$

Solve for $z$.
FOIL out both sides:
$\displaystyle (a + z)^2 = a^2 + 2az + z^2$

$\displaystyle \left ( \frac{a(2a + z)}{r} \right ) ^2 = \left ( \frac{a^2 (2a + z)^2}{r^2} \right )$

Can you finish from here?

-Dan

9. ## Re: Optimisation question

Originally Posted by topsquark
FOIL out both sides:
$\displaystyle (a + z)^2 = a^2 + 2az + z^2$

$\displaystyle \left ( \frac{a(2a + z)}{r} \right ) ^2 = \left ( \frac{a^2 (2a + z)^2}{r^2} \right )$

Can you finish from here?

-Dan
Hello,
So I got $r=a*\sqrt{2}$ and h=4*a and minimum volume of a cone is $\frac83*a^3*π$

10. ## Re: Optimisation question

Originally Posted by Vinod
Hello,
So I got $r=a*\sqrt{2}$ and h=4*a and minimum volume of a cone is $\frac83*a^3*π$
That's what I got as well.