The cylinder bore in the form of circular cylinder open at the top is to be made from a metal sheet of area 75πcm 
2
 .
Based on the above information, answer the following questions
:(i) If the radius of cylinder is rcm and height is hcm, then write the volume V of cylinder in terms of radius r.
(ii) Find  
dr
dV
​
 
(iii) (a) Find the radius of cylinder when its volume is maximum.

Given the problem involving a circular cylinder open at the top made from a metal sheet of area \( 75\pi \, \text{cm}^2 \), we can address the following parts step by step.

### Part (i): Writing the Volume \( V \) of the Cylinder in Terms of Radius \( r \)

The volume \( V \) of a cylinder is given by the formula:

\[
V = \pi r^2 h
\]

where:
- \( r \) is the radius of the base of the cylinder,
- \( h \) is the height of the cylinder,
- \( \pi \) is a constant.

### Part (ii): Finding \( \frac{dV}{dr} \)

To differentiate the volume function with respect to the radius \( r \), we first need to express the height \( h \) in terms of \( r \).

The surface area of the cylinder consists of two parts:
1. The lateral surface area (side) of the cylinder: \( 2\pi r h \),
2. The base area: \( \pi r^2 \) (only one base since the cylinder is open at the top).

Thus, the total surface area \( S \) of the cylinder is:

\[
S = \text{lateral surface area} + \text{base area} = 2\pi r h + \pi r^2
\]

Given that the surface area is \( 75\pi \, \text{cm}^2 \), we have:

\[
2\pi r h + \pi r^2 = 75\pi
\]

Dividing through by \( \pi \), we get:

\[
2rh + r^2 = 75
\]

Solve for \( h \):

\[
h = \frac{75 - r^2}{2r}
\]

Now substitute this expression for \( h \) into the volume formula \( V = \pi r^2 h \):

\[
V = \pi r^2 \left(\frac{75 - r^2}{2r}\right)
\]

Simplify the expression:

\[
V = \pi r^2 \times \frac{75 - r^2}{2r} = \frac{\pi r(75 - r^2)}{2}
\]

Thus, the volume in terms of \( r \) is:

\[
V = \frac{\pi}{2}(75r - r^3)
\]

To find \( \frac{dV}{dr} \), differentiate this expression with respect to \( r \):

\[
\frac{dV}{dr} = \frac{\pi}{2} \left( 75 - 3r^2 \right)
\]

### Part (iii): Finding the Radius for Maximum Volume

To find the radius when the volume is maximum, set \( \frac{dV}{dr} = 0 \):

\[
\frac{\pi}{2}(75 - 3r^2) = 0
\]

Solving for \( r \):

\[
75 - 3r^2 = 0
\]

\[
3r^2 = 75
\]

\[
r^2 = 25
\]

\[
r = 5 \, \text{cm}
\]

Thus, the radius of the cylinder when the volume is maximum is \( r = 5 \, \text{cm} \).

### Summary:

- **(i)** Volume of the cylinder in terms of radius: \( V = \frac{\pi}{2}(75r - r^3) \)
- **(ii)** The derivative \( \frac{dV}{dr} = \frac{\pi}{2}(75 - 3r^2) \)
- **(iii)** The radius of the cylinder when the volume is maximum: \( r = 5 \, \text{cm} \)

Would you like more details on any part of this solution?

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### Relative Questions:
1. How does the surface area constraint affect the volume of the cylinder?
2. What happens to the height when the radius increases?
3. How can we confirm that the radius of 5 cm gives a maximum volume?
4. How would the solution change if the cylinder were closed at the top?
5. What is the corresponding height when the volume is maximum?

### Tip:
Maximizing the volume of a cylinder with a fixed surface area often leads to a relationship between radius and height. Consider this relationship when dealing with such optimization problems.

The cylinder bore in the form of circular cylinder open at the top is to be made from a metal sheet of area 75π cm². Based on the above information, answer the following questions: (i) If the radius of cylinder is r cm and height is h cm, then write the volume V of cylinder in terms of radius r. (ii) Find dV/dr. (iii) (a) Find the radius of cylinder when its volume is maximum.

Solve a cylinder optimization problem where the volume is maximized given a surface area of 75π cm². Learn how to express volume in terms of radius, differentiate, and find the optimal radius for maximum volume. This problem involves geometry, surface area-volume relationships, and differentiation. Suitable for students in Grades 10-12.

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