Problem 41 - Single Variable Calculus

Question

Packaging A container of capacity 64 in. \(^{3}\) is to be made in the form of a right circular cylinder. The top and the bottom of the can are to be cut from squares, whereas the side is to be made by bending a rectangular sheet so that the ends match. Find the radius and height of the can that can be constructed with the least amount of material.

Step-by-Step Solution

Answer

The optimal dimensions of the container that minimize the material used are approximately a radius of \(r \approx 2\) inches and a height of \(h = \frac{16}{\pi}\) inches.

Step 1: Assign symbols to variables

Let's assign symbols to our variables: - Let the radius of the cylindrical container be r. - Let the height of the cylindrical container be h.

Step 2: Define given constraints and find relationships

The given constraint is that the volume of the cylindrical container is 64 in\(^3\). The volume of a cylinder is given by the formula: \[ V = \pi r^2 h \] We are given that V = 64 in\(^3\). So, we have the equation: \[ 64 = \pi r^2 h \] Now we need to find a relationship between r and h so that we can write the equation involving only one of the unknowns.

Step 3: Formulate expression for the material used

The container is made up of two square lids and a cylindrical side. The material used for the top and bottom is: \[ 2(r^2 + h^2) \] (Because both the top and bottom are cut from squares) The material used for the side is equal to the area of the rectangle formed when the cylindrical side is laid flat. This rectangle will have length equal to the circumference of the cylinder, and its width equal to h. So, the area of the rectangle is: \[ 2\pi rh \] Now, we can find the total amount of material used for constructing the container by adding these two quantities. \[ M(r, h) = 2(r^2 + h^2) + 2\pi rh \]

Step 4: Express material used as a function of a single variable

Now let's eliminate h by substituting from the volume equation: \[ h = \frac{64}{\pi r^2} \] Substitute h into the expression for the material used: \[ M(r) = 2(r^2 + (\frac{64}{\pi r^2})^2) + 2\pi r (\frac{64}{\pi r^2}) \]

Step 5: Find the minimum value of the material used

To find the minimum value of M(r), we need to find the critical points by finding the derivative of M(r) with respect to r, and then set it to 0.

Step 6: Calculate the derivative of M(r)

Calculate the derivative of M(r) with respect to r: \[ M'(r) = \frac{dM(r)}{dr} \] After evaluating the derivative, we have: \[ M'(r) = 4r - \frac{2048}{\pi r^3} + 64 \]

Step 7: Find critical points by setting M'(r) equal to 0

Now, we need to find the values of r for which M'(r) is equal to 0: \[ 4r - \frac{2048}{\pi r^3} + 64 = 0 \]

Step 8: Solve for r

Solving this equation for r, we get approximately: \[ r \approx 2 \]

Step 9: Find h using r

Now that we have found r, let's find the corresponding value of h using the volume equation: \[ h = \frac{64}{\pi (2)^2} \] \[ h = \frac{64}{4\pi} \] \[ h = \frac{16}{\pi} \]

Step 10: Conclude the optimal dimensions of the container

The dimensions that minimize the material used for the container are: Radius: r \(\approx\) 2 in Height: h = \(\frac{16}{\pi}\) in

Key Concepts

right circular cylinder

A right circular cylinder is a three-dimensional shape with two parallel, identical circular bases and a curved surface connecting these bases. It has a central axis perpendicular to the bases which gives it the 'right' descriptor. This type of cylinder is commonly used in real-world objects like cans. For mathematical purposes, we usually focus on its volume and surface area, which are functions of its radius and height. The volume is given by the formula \( V = \pi r^2h \). The surface area includes the areas of the two bases and the curved surface that forms the side, which is calculated using \( 2 \pi rh \). A deep understanding of these elements is crucial for solving optimization problems involving cylinders.

volume constraint

The volume constraint is a key component when dealing with optimization problems in cylinders. In this exercise, the volume is constrained at 64 in\(^3\). This means that our cylinder must hold exactly this amount of space. Using the formula for volume, \( V = \pi r^2 h \), we need to find a relationship between the radius \( r \) and the height \( h \) that holds this volume consistent while minimizing other factors like material use. This leads to creating an equation where \( r \) and \( h \) are dependent on each other, which then simplifies our equation into working with a single variable.

derivative for critical points

Finding critical points by using derivatives is important in calculus for optimization problems. Critical points occur where the derivative of a function is zero, indicating potential maximum and minimum points. To reduce material used in constructing the cylinder, the derivative of the materials function with respect to \( r \), \( M'(r) \), is calculated. This derivative is set to zero:\[4r - \frac{2048}{\pi r^3} + 64 = 0\]Solving this equation helps us determine the radius at which material use is minimized. Plugging this \( r \) back into the constraints will give associated values for \( h \), completing the dimensions that optimize the material usage.

material minimization

Concrete applications involve minimizing materials for cost or efficiency reasons. In the scenario where you need to construct a cylindrical can, material minimization means using as little material for the structure while maintaining its volume. The total material is calculated from the area of the two circular ends and the curved side. - Top and bottom are formed from squares, thus their contribution is \( 2(r^2 + h^2) \).- The side material forms a rectangle when uncurled, calculated as \( 2\pi rh \). Summing these gives \( M(r, h) \). By finding a way to express \( M \) in terms of a single variable, then finding its derivative, we identify the optimal \( r \) and \( h \) that satisfy both the volume constraint and minimum material use. This process demonstrates how calculus helps in creating solutions that are both efficient and economical.