Q1. [Maximum Marks: 6]
A cylindrical container is to be constructed with a closed base but no top. The container
must hold 1000 cubic centimeters of liquid.
(a) Write an expression for the surface area of the container in terms of the radius r of
the base.
(b) Use calculus to determine the dimensions of the container (radius anc height) that
minimize the surface area.
(c) What is the minimum surface area of the container?

Let's solve this step-by-step.

### (a) Expression for Surface Area
We are given that the cylindrical container has no top and a closed base, and it must hold 1000 cubic centimeters of liquid.

Let:
- \( r \) be the radius of the base of the cylinder.
- \( h \) be the height of the cylinder.

The volume of a cylinder is given by:
\[
V = \pi r^2 h
\]
We are told that the volume must be 1000 cm³:
\[
\pi r^2 h = 1000
\]
From this, solve for \( h \):
\[
h = \frac{1000}{\pi r^2}
\]

The surface area \( A \) of the cylinder (without the top) consists of the area of the base and the lateral surface area:
\[
A = \text{Area of base} + \text{Lateral surface area}
\]
\[
A = \pi r^2 + 2 \pi r h
\]
Substitute \( h = \frac{1000}{\pi r^2} \) into this equation:
\[
A = \pi r^2 + 2 \pi r \left( \frac{1000}{\pi r^2} \right)
\]
Simplify:
\[
A = \pi r^2 + \frac{2000}{r}
\]

Thus, the expression for the surface area in terms of \( r \) is:
\[
A(r) = \pi r^2 + \frac{2000}{r}
\]

### (b) Minimizing the Surface Area
To minimize the surface area, we need to find the critical points of \( A(r) \). This involves taking the derivative of \( A(r) \) and setting it equal to zero.

First, compute the derivative of \( A(r) \):
\[
A'(r) = \frac{d}{dr} \left( \pi r^2 + \frac{2000}{r} \right)
\]
\[
A'(r) = 2 \pi r - \frac{2000}{r^2}
\]
Set the derivative equal to zero to find critical points:
\[
2 \pi r - \frac{2000}{r^2} = 0
\]
Rearrange:
\[
2 \pi r = \frac{2000}{r^2}
\]
Multiply both sides by \( r^2 \):
\[
2 \pi r^3 = 2000
\]
Solve for \( r \):
\[
r^3 = \frac{2000}{2 \pi} = \frac{1000}{\pi}
\]
\[
r = \left( \frac{1000}{\pi} \right)^{1/3}
\]
This simplifies to approximately:
\[
r \approx 6.83 \text{ cm}
\]

Now, use the volume equation to find \( h \). Since \( h = \frac{1000}{\pi r^2} \):
\[
h = \frac{1000}{\pi (6.83)^2} \approx 6.83 \text{ cm}
\]

Thus, the dimensions of the container that minimize the surface area are:
- Radius \( r \approx 6.83 \) cm
- Height \( h \approx 6.83 \) cm

### (c) Minimum Surface Area
To find the minimum surface area, substitute \( r \approx 6.83 \) cm into the surface area formula:
\[
A(r) = \pi r^2 + \frac{2000}{r}
\]
\[
A(6.83) = \pi (6.83)^2 + \frac{2000}{6.83}
\]
\[
A \approx 146.97 + 292.92 \approx 439.89 \text{ cm}^2
\]

Thus, the minimum surface area of the container is approximately:
\[
\boxed{439.89 \text{ cm}^2}
\]

---

Would you like more details or have any questions? 

Here are 5 related questions:
1. How does the inclusion of a top affect the surface area optimization?
2. Can this method be applied to containers with other shapes, such as cones?
3. What is the significance of the derivative in finding the minimum value?
4. How do constraints like volume or cost affect such optimization problems?
5. How does changing the volume to a different value affect the dimensions and surface area?

**Tip:** When solving optimization problems, always check that the second derivative confirms whether it's a minimum or maximum.

Q1. [Maximum Marks: 6] A cylindrical container is to be constructed with a closed base but no top. The container must hold 1000 cubic centimeters of liquid. (a) Write an expression for the surface area of the container in terms of the radius r of the base. (b) Use calculus to determine the dimensions of the container (radius and height) that minimize the surface area. (c) What is the minimum surface area of the container?

Explore how to minimize the surface area of a cylindrical container with no top while maintaining a fixed volume of 1000 cubic centimeters. This problem uses calculus and optimization concepts, with key formulas for surface area and volume. Suitable for students in Grades 11-12.

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