# Maxima and Minima in calculus

• MHB
• WMDhamnekar
In summary, the radius of the right circular cylinder of greatest curved surface area inscribed in a given cone is half of the cone's radius. This can be proven by setting up a coordinate system and using the formula for the curved surface area of a cylinder. Taking the derivative and setting it equal to zero, we can find the maximum value for r which happens to be half of the cone's radius. A picture can also help illustrate this concept. Additionally, using a similar coordinate system, we can prove that the ratio of the cylinder's height to the cone's height is equal to the ratio of the difference between the cone's radius and the cylinder's radius to the cone's radius.
WMDhamnekar
MHB
Question: Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone.

Let r and h be the radius and height of the right circular cylinder inscribed in a given cone of radius R and height H. Let S be the curved surface area of cylinder.

S = 2πr*h

h = H*(R – r)/R
( Would any Math help board member provide me the detailed explanation of the computation of height of right circular cylinder of greatest curved surface inscribed in a given cone with a figure (as far as possible) ?

So S = 2πr*H(R – r)/R

= $\frac{2πH}{R}(r*R – r^2)$

Differentiate w.r.t.r

$\frac{dS}{dr} = \frac{2πH}{R}(R – 2r)$

For maxima or minima

$\frac{dS}{dr} =0$

=> $\frac{2πH}{R}(R – 2r) = 0$

=> R – 2r = 0

=> R = 2r

=> $r = \frac{R}{2}$

$\frac{d^2S}{dr^2} = \frac{2πH}{R}*(0 – 2)= \frac{-4πH}{R }$(negative)

So for $r = \frac{R}{2},$ S is maximum.

Draw a picture. From the side, a cone of radius R and height h looks like s triangle. I would set up a coordinate system with x-axis along the base, y-axis along the altitude, and origin at the center of the base. Then the peak is at (0, h) and one vertex is at (R, 0). The line between those two points, on the side of the cone, is given by y= -(h/R)x+ h. At x= r, y= -hr/R+ h= h(1- r/R).

The area of the curved side is $2\pi rh(1- r/R)$.

I drew a picture describing this question. Now, how can we prove $\frac{h}{H}=\frac{(R-r)}{R}$

Last edited:
The cone has height H and radius R. Set up a coordinate system so the origin is at the center of the base and the z axis passes through the vertex. Then the vertex is at (0, 0, H) and the x-axis passes through the cone at (R, 0, 0). The line through those two points, in the xz-plane, is given by $z= H\frac{R- x}{R}$.

Taking x= r, for the cylinder, we get $h= H\frac{R- r}{R}$ or, dividing both sides by H, $\frac{h}{H}= \frac{R- r}{R}$.

## 1. What are maxima and minima in calculus?

Maxima and minima are the highest and lowest values of a function, respectively. In calculus, they represent the points where the slope of a function changes from positive to negative (maxima) or negative to positive (minima).

## 2. How can I find the maxima and minima of a function?

To find the maxima and minima of a function, you can use the first or second derivative test. The first derivative test involves finding the critical points of the function, where the derivative is equal to zero or undefined. The second derivative test involves analyzing the concavity of the function at the critical points to determine if they are maxima or minima.

## 3. Why are maxima and minima important in calculus?

Maxima and minima are important in calculus because they help us understand the behavior of a function. They can be used to optimize a function, such as finding the maximum profit or minimum cost in a business setting. They also allow us to analyze the rate of change of a function and make predictions about its future behavior.

## 4. Can a function have more than one maxima or minima?

Yes, a function can have multiple maxima and minima. This can occur when the function has multiple critical points or when the function has a flat portion where the derivative is equal to zero.

## 5. How are maxima and minima related to the derivative of a function?

The derivative of a function represents the rate of change of the function. At maxima and minima, the derivative is equal to zero, indicating that the function is not changing at those points. This is why the first derivative test is used to find maxima and minima.

• Calculus
Replies
10
Views
2K
• Calculus
Replies
4
Views
337
• Calculus
Replies
2
Views
2K
• Calculus
Replies
2
Views
2K
• Calculus
Replies
2
Views
1K
• Calculus
Replies
20
Views
2K
• Mechanics
Replies
3
Views
487
• Introductory Physics Homework Help
Replies
3
Views
105
• Differential Geometry
Replies
7
Views
788
• Special and General Relativity
Replies
11
Views
164