Question

You are designing a soft drink container that has the shape of a right circular cylinder. The container is supposed to hold 12 fluid ounces (1 fluid ounce is approximately \(1.80469\) cubic inches). Find the dimensions that will use a minimum amount of construction material.

Short answer

The dimensions that will use a minimum amount of construction material are a radius of approximately 1.083 inches and a height of approximately 5.819 inches.

Step-by-step solution

Express the volume in terms of variables

We denote the radius of the cylinder as \( r \) and the height as \( h \). The volume \( V \) of a cylinder is given by the formula \( V = \pi r^2 h \), and we know from the problem that \( V = 12 \) fluid ounces, which is approximately \( 21.65628 \) cubic inches. Therefore, we have the equation \( \pi r^2 h = 21.65628 \). We solve for \( h \) to express \( h \) in terms of \( r \), yielding \( h = \frac{21.65628}{\pi r^2} \).

Express the surface area in terms of one variable

The surface area \( S \) of a cylinder is given by the formula \( S = 2\pi r(r + h) \). Substitute \( h = \frac{21.65628}{\pi r^2} \) from Step 1 into this formula to get \( S(r) = 2\pi r(r + \frac{21.65628}{\pi r^2}) \). We end up with \( S(r) = 2\pi r^2 + \frac{43.31256}{r} \). Now the surface area is written in terms of one variable \( r \).

Differentiate the surface area equation

Differentiate \( S(r) = 2\pi r^2 + \frac{43.31256}{r} \) with respect to \( r \) to get \( S'(r) = 4\pi r - \frac{43.31256}{r^2} \). This will be used to find the value of \( r \) that minimizes the surface area.

Find the value of r that minimizes surface area

Set \( S'(r) = 0 \) and solve for \( r \). This gives us \( r^3 = \frac{43.31256}{4\pi} \). Simplify to find that \( r \approx 1.083 \) inches. Substitute \( r = 1.083 \) inches into \( h = \frac{21.65628}{\pi r^2} \) to find \( h \approx 5.819 \) inches.

Key concepts

Cylinder Volume

The volume of a cylinder is a foundational concept in geometry. A cylinder is a 3D shape with two parallel circles (one at each end) and a curved surface connecting them. The volume of a cylinder helps us know how much space is inside it.
To find the volume, we use the formula:

• \( V = \pi r^2 h \)

Where:

• \( V \) is the volume
• \( r \) is the radius of the circular base
• \( h \) is the height of the cylinder

In our specific problem, we're working with a cylinder that holds 12 fluid ounces. Since 1 fluid ounce is approximately 1.80469 cubic inches, we multiply to find the total volume in cubic inches:\( V = 12 \times 1.80469 \approx 21.65628 \) cubic inches.
This result is crucial for further calculations, as it allows us to relate the radius and height through this volume constraint.

Surface Area of a Cylinder

Understanding the surface area of a cylinder is important, especially in design and manufacturing. The surface area is how much material we need to build the cylinder. The formula to calculate it is:

• \( S = 2\pi r(r + h) \)

This breaks down into:

• \( 2\pi r^2 \): the area of the two circular bases
• \( 2\pi rh \): the area of the curved surface

In our problem, we need to minimise the amount of construction material, meaning we want the smallest possible surface area.
By substituting the expression for height from the volume equation, \( h = \frac{21.65628}{\pi r^2} \), into the surface area formula, we express \( S \) as a function of \( r \):
\( S(r) = 2\pi r^2 + \frac{43.31256}{r} \).
This formula is now ready for optimization using calculus.

Derivatives in Calculus

Derivatives are a powerful tool in calculus that measure how a function changes as its input changes. In optimization problems like our cylinder design, they help us find points where a function reaches a minimum or maximum.
For our problem, we use the derivative of the surface area function, \( S(r) \), to find the radius \( r \) that minimizes the material used. The derivative, \( S'(r) \), is found by differentiating \( S(r) = 2\pi r^2 + \frac{43.31256}{r} \):

• \( S'(r) = 4\pi r - \frac{43.31256}{r^2} \)

To minimize the surface area, we set the derivative equal to zero and solve for \( r \):
\( 4\pi r - \frac{43.31256}{r^2} = 0 \).
Solving, we find \( r^3 = \frac{43.31256}{4\pi} \), which gives \( r \approx 1.083 \) inches.
This value is substituted back into the equation for height to find \( h \approx 5.819 \) inches, leading to the optimal dimensions for our container.