Problem 8 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

If an open box has a square base and a volume of \(108 \mathrm{in} .{ }^{3}\) and is constructed from a tin sheet, find the dimensions of the box, assuming a minimum amount of material is used in its construction.

Step-by-Step Solution

Answer

The dimensions of the open box with a minimal amount of material used are \(6\,\text{in} \times 6\,\text{in} \times 3\,\text{in}\), which are the side length of the square base and the height of the box, respectively.

Step 1: Express the volume and surface area as functions

Let \(x\) be the side length of the square base, and let \(h\) be the height of the box. Then the volume of the box is given by: \[V = x^2h\] We are given that the volume must be \(108\,\text{in}^3\), so we have: \[x^2h = 108\] The surface area of the box is given by the sum of the surface areas of the base and the four side walls. Since the box is open (no top), the surface area (A) can be expressed as: \[A = x^2 + 4(xh)\]

Step 2: Express h in terms of x

To solve this optimization problem, we need to express the surface area as a function of one variable. Using the volume constraint, we can rewrite the height \(h\) in terms of side length \(x\): \[h = \frac{108}{x^2}\]

Step 3: Replace h in the surface area equation

Now substitute the expression of \(h\) in the surface area function: \[A(x) = x^2 + 4\left(x\frac{108}{x^2}\right)\] Simplify to get: \[A(x) = x^2 + \frac{432}{x}\]

Step 4: Find the critical points

To find the minimum surface area, we need to find the critical points (where the derivative is zero). First, find the derivative of A(x) with respect to x: \[A'(x) = 2x - \frac{432}{x^2}\] Now, set the derivative equal to 0 and solve for x: \[2x - \frac{432}{x^2} = 0\] Add \(\frac{432}{x^2}\) to both sides: \[2x = \frac{432}{x^2}\] ⟹ \(x^3 = 216\) Take the cube root of both sides: \[x = 6\]

Step 5: Determine the values of x and h

We have found the critical point \(x=6\). Now, we need to find the corresponding value of \(h\). Use the volume constraint equation: \[h = \frac{108}{x^2}\] Substitute x=6 into the equation: \[h = \frac{108}{6^2} = \frac{108}{36} = 3\]

Step 6: Conclusion

The dimensions of the open box with a minimal amount of material used are 6 in × 6 in × 3 in, which are the side length of the square base and the height of the box, respectively.

Key Concepts

Calculus

Calculus is an essential tool for solving optimization problems, like finding the dimensions of a box that minimize surface area while maintaining a given volume. In calculus, we study how things change, focusing on the concepts of integration and differentiation. Here, differentiation is key because it allows us to calculate how the surface area of the box changes with changes in its dimensions. By setting the derivative of the surface area function to zero, we can locate the critical points, which indicate where minimum or maximum surface areas occur.

In our problem, the surface area of the open box is expressed as a function of the side length of its base. After finding the derivative of this function, we solve for the base side length that makes the derivative zero, indicating a potential minimum surface area. Calculus provides the mathematical groundwork to systematically address such real-world modeling problems by precisely analyzing rates of change and behavior of functions.

Surface Area

The surface area of a shape is the total area covered by its outer surfaces. For an open box, like in this exercise, the surface area comprises the area of the base plus the areas of the four sides. Since it's open, there is no top. Mathematically, we can express the surface area as:

• Base area: \(x^2\) square inches.
• Side areas: Four times the product of base side length \(x\) and height \(h\).

Thus, the total surface area \(A\) is given by: \[A = x^2 + 4xh\]In this exercise, minimizing the amount of material used for the box translates to minimizing the surface area function. We manipulate this expression using a constraint (the fixed volume) to find the optimal dimensions. The calculus-derived critical points determine which exact dimensions help achieve the minimal material use.

Volume Constraint

In optimization problems, constraints often define relationships that must be maintained while optimizing a desired objective. Here, the volume constraint is that the box must hold 108 cubic inches. This constraint creates a relationship between the height \(h\) and the base side length \(x\) of the box:\[V = x^2 h = 108 \]
By solving this equation for \(h\), we express the height in terms of the other variable \(x\), which helps in reducing the surface area function to a single variable. This reduction is crucial for the optimization process, as it simplifies the task of finding critical points where the minimum surface area occurs.

With the formula \(h = \frac{108}{x^2}\), we successfully incorporate the volume constraint into the overall optimization, ensuring that while we aim for minimal surface area, the box still maintains the specified volume of 108 cubic inches. Constraints like these make real-world problems challenging yet more structured and mathematically intriguing.