# Maximum area for inscribed cylinder

• Elias Waranoi
In summary: Setting... In summary, the result of the derivation is that there is no maximum or minimum in the growth of area for a cylinder inscribed in a cone.
Elias Waranoi

## Homework Statement

Inscribe in a given cone, the height h of which is equal to the radius r of the base, a cylinder (c) whose total area is a maximum. Radius of cylinder is rc and height of cylinder is hc.

## Homework Equations

A = 2πrchc + 2πrc2

## The Attempt at a Solution

r = h ∴ hc = r - rc
A = 2πrc(r - rc) + 2πrc2

To get the maximum of this area I will find the radius rc when the growth of the area is zero.
dA/drc = 0 = 2πr

What does this result mean? I don't understand how 0 = 2πr makes sense as a result from a derivation. What kind of information does this result give me geometrically? How can I know that there is no maximum area to the cylinder as my answer sheet tells me.

 oops, I checked my result and now we agree. Looks as if the area does not depend on rc -- as I could have read in your post

 oops2 Thanks Ray for putting me right -- dA/drc does not depend on rc but the area itself of course does. It just keeps increasing until rc hits r and then you have a pancake, not a cylinder any more. Tricky exercise if you are doing it before coffee

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Elias Waranoi said:

## Homework Statement

Inscribe in a given cone, the height h of which is equal to the radius r of the base, a cylinder (c) whose total area is a maximum. Radius of cylinder is rc and height of cylinder is hc.

## Homework Equations

A = 2πrchc + 2πrc2

## The Attempt at a Solution

r = h ∴ hc = r - rc
A = 2πrc(r - rc) + 2πrc2

To get the maximum of this area I will find the radius rc when the growth of the area is zero.
dA/drc = 0 = 2πr

What does this result mean? I don't understand how 0 = 2πr makes sense as a result from a derivation. What kind of information does this result give me geometrically? How can I know that there is no maximum area to the cylinder as my answer sheet tells me.

Your problem is constrained: ##\max f = r_c h_c + r_c^2## subject to ##r_c+h_c = r##. The objective is ##A/(2 \pi)## and the constraint arises because the cylinder is inscribed in a cone of base-radius and height ##r##. You can substitute ##h_c = r - r_c## into ##f## to get a one-dimensional problem ##\max f_1 = r_c^2 + r_c(r-r_c) = r r_c,## subject to ##0 \leq r_c \leq r##. What is the solution to this last problem? (Remember that ##r## is a constant, so ##r_c## is the only variable.)

Ray Vickson said:
Your problem is constrained: maxf=rchc+r2cmaxf=rchc+rc2\max f = r_c h_c + r_c^2 subject to rc+hc=rrc+hc=rr_c+h_c = r. The objective is A/(2π)A/(2π)A/(2 \pi) and the constraint arises because the cylinder is inscribed in a cone of base-radius and height rrr. You can substitute hc=r−rchc=r−rch_c = r - r_c into fff to get a one-dimensional problem maxf1=r2c+rc(r−rc)=rrc,maxf1=rc2+rc(r−rc)=rrc,\max f_1 = r_c^2 + r_c(r-r_c) = r r_c, subject to 0≤rc≤r0≤rc≤r0 \leq r_c \leq r. What is the solution to this last problem? (Remember that rrr is a constant, so rcrcr_c is the only variable.)

Sorry, bad title. What I'm looking for is the maximum in the curve of the growth of area for a cylinder inscribed in a cone. Not the maximum area. What I'm hung up on the result of my derivation dA/drc = 0 = 2πr. If for example my cone had the radius r of 1 meter than my derivation would give dA/drc = 2πr = 6.28 = 0 which doesn't make sense. But dA/drc = 0 is just a statement I made and if dA/drc ≠ 0 then that means my statement is false and there is no maximum or minimum in the growth of area for the cylinder. Seems like I figured it out while writing me reply... Sorry for wasting your time and thank you all for trying to help me!

Elias Waranoi said:
Sorry, bad title. What I'm looking for is the maximum in the curve of the growth of area for a cylinder inscribed in a cone. Not the maximum area. What I'm hung up on the result of my derivation dA/drc = 0 = 2πr. If for example my cone had the radius r of 1 meter than my derivation would give dA/drc = 2πr = 6.28 = 0 which doesn't make sense. But dA/drc = 0 is just a statement I made and if dA/drc ≠ 0 then that means my statement is false and there is no maximum or minimum in the growth of area for the cylinder. Seems like I figured it out while writing me reply... Sorry for wasting your time and thank you all for trying to help me!

Setting the derivative to zero is a mistake: you have a constrained problem! That is, your originally-described problem is
$$\begin{array}{l}\max A(r_c) = 2 \pi r_c(r-r_c) + 2 \pi r_c^2 = k r_c, \\ \text{subject to} \;\;0 \leq r_c \leq r. \end{array}$$
Here, ##k = 2 \pi r## is a positive constant.

Your newly-described problem is to maximize the constant ##k## on the set ##0 \leq r_c \leq r##, which makes no sense: the area just keeps growing at a steady rate, so every point is a maximum growth rate point.

Setting a derivative to zero is what you do when you are looking for an interior-point max or min of a function, but not when you are on the boundary of an inequality constraint set. You would never, ever solve ##\max/ \min f(x) = 2x, \: 0 \leq x \leq 1## by setting the derivative of ##2x## to zero.

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## 1. What is the formula for finding the maximum area of an inscribed cylinder?

The formula for finding the maximum area of an inscribed cylinder is A = πr2h, where r is the radius of the base and h is the height of the cylinder.

## 2. How do you determine the maximum area of an inscribed cylinder?

To determine the maximum area of an inscribed cylinder, you need to find the dimensions that will result in the largest possible value for A, using the formula A = πr2h. This can be done by using calculus and finding the critical points of the function or by using geometric reasoning and the fact that the maximum area will occur when the cylinder is tangent to the base of the larger shape it is inscribed in.

## 3. What is the relationship between the maximum area of an inscribed cylinder and the shape it is inscribed in?

The maximum area of an inscribed cylinder is directly related to the shape it is inscribed in. The larger the base shape, the larger the maximum area of the inscribed cylinder can be. The shape and dimensions of the base shape will also affect the dimensions of the inscribed cylinder and therefore its maximum area.

## 4. Can the maximum area of an inscribed cylinder be greater than the area of the base shape?

No, the maximum area of an inscribed cylinder cannot be greater than the area of the base shape. This is because the cylinder is inscribed in the base shape, meaning that all of its points must lie within the boundaries of the base shape. Therefore, the area of the cylinder will always be less than or equal to the area of the base shape.

## 5. What is the significance of finding the maximum area of an inscribed cylinder?

Finding the maximum area of an inscribed cylinder is important in many practical applications, such as maximizing storage space in containers or optimizing the design of cylindrical structures. It also helps develop problem-solving skills and a deeper understanding of mathematical concepts, such as optimization and geometric relationships.

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