Problem 18 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Q-Tips \(^{\otimes}\) are a brand of cotton swabs each 3 inches long. You can purchase a pack of 300 of them in a plastic rectangular container backed in cardboard. In other words, the plastic forms an open box and the "lid" is cardboard. The width of the box is 3 inches. What should the length and depth be if the goal is to minimize the amount of plastic used? In order to hold 300 Q-Tips the box must have a volume of \(33.75\) square inches. In reality, such a box is \(7.5\) inches long and \(1.5\) inches deep. Has the amount of plastic been minimized?

Step-by-Step Solution

Answer

The amount of plastic is not minimized in the given box dimensions. The optimal box dimensions that would minimize plastic use with the given width and volume are approximately length of 2.24 inches and depth of 5.02 inches.

Step 1: Set Up the Constraints

In this case, the constraint is the volume of the box. The product of the length \(l\), width \(w\) and depth \(d\) - which forms the box's volume - must equal 33.75 cubic inches. So we have the equation: \(l*w*d = 33.75\). Given that the width is 3 inches, we can rework this equation into: \(l*d = 11.25\). This equation tells us the product of the depth and length and we can use it to express \(d\) in function of \(l\) as \(d = 11.25/l\).

Step 2: Set Up the Function to be Minimized

The function to minimize is the surface area of the box. The formula for that is: 2*(length*width + width*depth + depth*length). Since we have the width \(w = 3\) inches, we can substitute \(w\) and \(d\) into the function. We write the surface area \(S\) as a function of \(l\), where \(S(l) = 2 * (3*l + 3*(11.25/l) + (11.25/l)*3)\). Now, the task is to find the length \(l\) which minimizes this function.

Step 3: Find the Minimum

To find the minimum of \(S(l)\), we can differentiate it and find where the derivative equals 0. Taking the derivative, we get: \( S'(l) = 2*3 - 2 * (3 * (11.25 / l^2)). By setting this equal to 0 and solving for \(l\), we find that the minimum occurs when \(l = \sqrt[3]{11.25} \approx 2.24\) inches

Step 4: Verification

If we substitute these values of \(l = 2.24\) and \(w = 3\) into our equation for \(d = 11.25/l\), we find that \(d = 5.02\) inches. Now we can compare these values to the dimensions present on the real box, \(l=7.5 \) and \(d=1.5\). It seems that, given the constraints of the problem, the real box does not minimize the amount of plastic being used.

Key Concepts

Understanding Volume Constraints

When dealing with real-world packaging problems, optimizing the amount of material used is crucial for both environmental sustainability and cost-effectiveness. Volume constraints play a significant role in these optimization problems. In calculus, a constraint is a condition that must be met within a problem. For the Q-Tips packaging, the volume constraint is that the plastic box must hold 300 Q-Tips, translating into a fixed volume of 33.75 cubic inches.

Once this constraint is set, it guides how the dimensions of the box are chosen. In our problem, the width is pre-determined to be 3 inches, necessitating that the length and depth must be adjusted accordingly to achieve the desired volume. Using the formula for the volume of a rectangular prism, which is the product of its length, width, and depth (\( V = l \times w \times d \)), we can derive relationships between these dimensions that adhere to the volume restriction.

Function Minimization in Packaging

Function minimization is a powerful concept in calculus often applied to various optimization problems, including packaging design like our Q-Tips box example. The objective here is to minimize the amount of plastic used, which can be modeled mathematically by a function that represents the surface area of the plastic container.

In the context of the Q-Tips box, the function we want to minimize is the surface area formula, adjusted for our specific dimensions and constraints. To achieve the minimum surface area—thus using the least amount of plastic—we express the surface area as a function of one variable (\( l \) or \( d \) in this case) using the volume constraint. Afterward, we can use tools in calculus to find the minimum value of this function.

Derivatives in Calculus for Optimization

The derivative is a fundamental tool in calculus used to pinpoint rates of change and slope values of functions. When applied to optimization problems, the first derivative test allows us to find local minima and maxima of functions, which correspond to the least or the greatest values the function can take, respectively.

In our problem, after expressing the surface area as a function of one variable (\( S(l) \font\), we differentiate with respect to that variable to find its derivative (\( S'(l) \font\). Setting this derivative equal to zero provides us with critical points, which are candidates for our minimization. Solving this equation helps us find the optimal length (\( l \font\), and by extension, the optimal depth (\( d = 11.25/l \font\), to minimize the surface area of the box.

Surface Area Calculation to Minimize Material Usage

The surface area of a rectangular box is the sum of the areas of all six faces. For an open box like the Q-Tips container, which lacks a top, we calculate the area of only five faces. Efficiently utilizing material dictates calculating the minimum surface area required to make a box that fits the volume constraints.

In the exercise, the formula for surface area is 2*(length*width + width*depth + depth*length). Given that the width does not change, the challenge is to alter the length and depth to minimize this value. By expressing depth in terms of length using the volume constraint (\( d = 11.25/l \font\), we simplify the problem to a single variable and proceed to find the optimal solution using derivatives, as described in the previous sections. This process is vital for ensuring material optimization in packaging designs.