# Thread: Optamizing Lateral Surface Area of a Cylinder

1. ## Optamizing Lateral Surface Area of a Cylinder

A right circular cone has a base radius 5 and altitude 12. A cylinder is to be inscribed in the cone so that the axis of the cylinder coincides with the axis of the cone. Given that the radius of the cylinder must be between 2 and 4 inclusive, find the value of that radius for which the lateral surface area of the cylinder in minimum. Justify your answer. (Note: The lateral surface of a cylinder does NOT include the bases.)

Again I'm not good with Calculus word problems. Am I suppose to find the derivative of the lateral surface area of the cylinder? How do I handle the variables r and altitude? Help please. I really tried to get it, but I'm not use to them.

2. Originally Posted by ment2byours
A right circular cone has a base radius 5 and altitude 12. A cylinder is to be inscribed in the cone so that the axis of the cylinder coincides with the axis of the cone. Given that the radius of the cylinder must be between 2 and 4 inclusive, find the value of that radius for which the lateral surface area of the cylinder in minimum. Justify your answer. (Note: The lateral surface of a cylinder does NOT include the bases.)

Again I'm not good with Calculus word problems. Am I suppose to find the derivative of the lateral surface area of the cylinder? How do I handle the variables r and altitude? Help please. I really tried to get it, but I'm not use to them.

consider $a=2 \pi rh$

$h=12-\frac{12}{5}*r$

for volume we get $a=2\pi r(12-\frac{12r}{5}$

set it to zero find the derivative

$\frac{da}{dr}=24\pi-\frac{48}{5}\pi r$

$\frac{48}{5} \pi r=24 \pi$

$\boxed{r=\frac{5}{2}}$

3. Can you please show me how did you get h? I know from r/h=5/12 you get h=12/5, but not the rest. thanks again!

4. Originally Posted by choi_siwon
I'm having trouble on how to do this problem or where to go on this... I know what the picture would look like and that's about it. Any help would be really appreciated! Thank you.

A right circular cone has base radius 5 and altitude 12. A cylinder is to be inscribed in the cone so that the axis the cylinder coincides with the axis of the cone. Given that the radius of the cylinder must be between 2 and 4 inclusive, find the value of that radius for which the lateral surface area of the cylinder is minimum. Justify your answer and note that the lateral surface of a cylinder does NOT include the bases.
Draw a side-on diagram. Let the radius and height of the cylinder be r and h respectively.

From similar triangles: $\frac{5 - r}{h} = \frac{5}{12}$ .... (1)

Lateral surface area of cylinder: $S = \pi r^2 h$ .... (2)

Use (1) to express $S$ as a function of $r$ only.

Use calculus to find the value of $r$ that minimises $S$.

5. Draw the side view, being an isosceles triangle with a rectangle inside it.

Note that the rectangle, in touching the sides of the triangle, cuts off smaller triangles which must be similar to half of the original triangle.

Note that, whatever the value of the cylinder's radius "r" is, the base of the lower small triangle (let's look at the one on the right of the rectangle) is 5 - r. Also, the height "h" of this smaller triangle is related, by similarity, to the height of the original triangle by:

. . . . . $\frac{h}{12}\, =\, \frac{5\, -\, r}{5}$

Solve this relation to get "h" in terms only of "r", noting that the height of the smaller triangle is also the height of the cylinder. Then try to create a formula for the surface area of the "sides" of the cylinder in terms only of "r", and optimize. Remember to check the endpoints of the given interval.

6. Originally Posted by ment2byours
Can you please show me how did you get h? I know from r/h=5/12 you get h=12/5, but not the rest. thanks again!
See here: http://www.mathhelpforum.com/math-he...ibed-cone.html

7. Originally Posted by TheMasterMind
consider $a=2 \pi rh$

$h=12-\frac{12}{5}*r$

for volume we get $a=2\pi r(12-\frac{12r}{5}$

set it to zero find the derivative

$\frac{da}{dr}=24\pi-\frac{48}{5}\pi r$

$\frac{48}{5} \pi r=24 \pi$

$\boxed{r=\frac{5}{2}}$
That makes sense, but the only thing I dont understand is...how do you know whether r=2.5 is a max or a min?

8. Originally Posted by alakaboom1
That makes sense, but the only thing I dont understand is...how do you know whether r=2.5 is a max or a min?
note that A is quadratic in r with a leading coefficient < 0 ... graph is a parabola that opens down, indicating a max.

or

note that $\frac{d^2A}{dr^2} < 0$ for all r.

9. Originally Posted by skeeter
note that A is quadratic in r with a leading coefficient < 0 ... graph is a parabola that opens down, indicating a max.

or

note that $\frac{d^2A}{dr^2} < 0$ for all r.
I see, so 2.5 is the maximum radius. So, the minimum radius would be one of the 2 endpoints?

10. Originally Posted by alakaboom1
I see, so 2.5 is the maximum radius. So, the minimum radius would be one of the 2 endpoints?
no ... maximum area occurs when r = 2.5

endpoints would indicate values of r where area is a minimum.

11. Originally Posted by skeeter
no ... maximum area occurs when r = 2.5

endpoints would indicate values of r where area is a minimum.
sorry, thats what I meant to say. So, since the problem asks for the radius in which the area will be minimum, the answer will be one of the 2 endpoints. Cool, thanks a lot.

12. Originally Posted by ment2byours
Can you please show me how did you get h? I know from r/h=5/12 you get h=12/5, but not the rest. thanks again!
i apologize for the lack of explanation, i will try to be more thorough next time