Problem 26 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

An open rectangular box having a volume of 108 in. \(^{3}\) is to be constructed from a tin sheet. Find the dimensions of such a box if the amount of material used in its construction is to be minimal. Hint: Let the dimensions of the box be \(x^{\prime \prime}\) by \(y^{\prime \prime}\) by \(z^{\prime \prime}\). Then, \(x y z=108\) and the amount of material used is given by \(S=x y+\) \(2 y z+2 x z .\) Show that $$S=f(x, y)=x y+\frac{216}{x}+\frac{216}{y}$$ Minimize \(f(x, y)\)

Step-by-Step Solution

Answer

The dimensions of the open rectangular box that minimize the amount of material used are approximately \(x \approx 3\), \(y \approx 6\), and \(z \approx 6\) inches.

Step 1: Rewrite the volume equation

As given, the volume equation for the box is \(xyz = 108\). To express z in terms of x and y, we can rewrite this equation as: \[z = \frac{108}{xy}\]

Step 2: Substitute the expression for z into the equation for f(x, y)

Now that we have z in terms of x and y, we can substitute it into the surface area function to eliminate z. The given equation for the amount of material used is: \[S = xy + 2yz + 2xz\] Substituting the expression for z from Step 1, we get: \[S = xy + 2y\left(\frac{108}{xy}\right) + 2x\left(\frac{108}{xy}\right)\] Simplifying the equation: \[S = xy + \frac{216}{x} + \frac{216}{y}\] Our function becomes: \[f(x, y) = xy + \frac{216}{x} + \frac{216}{y}\]

Step 3: Find the critical points of f(x, y)

To find the critical points of the function f(x, y), we need to find the partial derivatives with respect to both x and y, and then set them equal to 0 and solve the system of equations: \[\frac{\partial f(x, y)}{\partial x} = y - \frac{216}{x^2}\] \[\frac{\partial f(x, y)}{\partial y} = x - \frac{216}{y^2}\] Setting both partial derivative equations to 0: \[y - \frac{216}{x^2} = 0\] \[x - \frac{216}{y^2} = 0\] Solve the system of equations above to find x and y, which will give us the critical points.

Step 4: Determine the dimensions that minimize the surface area

The critical points found will give us the dimensions of the box that minimize the surface area. From the critical points, we can also use the volume equation from Step 1 to find the corresponding z value, finally providing us with the complete dimensions x, y, and z that minimize the amount of material used in the construction of the open rectangular box.

Key Concepts

Critical Points

In calculus, critical points are essential in understanding the behavior of a function. These points are where the first derivative of the function equals zero or is undefined. In the context of optimization problems like this, we seek to find critical points because they help identify potential locations of minimum or maximum values, which are crucial for problems that ask us to minimize or maximize a quantity.
For the function of surface area, we calculate the partial derivatives with respect to each variable. Then we set these derivatives equal to zero. This process gives us the critical points. It's like finding a turning point on a curve where the slope is flat or undefined. Here, such points indicate potential box dimensions that may minimize material usage, making it a practical application of critical points in real-world scenarios.

Surface Area Minimization

Surface area minimization involves finding the smallest possible surface area for a given structural requirement, such as volume. In this exercise, we aim to minimize the surface area using a given volume constraint of 108 cubic inches. This optimization not only reduces the material needed, ultimately lowering cost, but also supports sustainability by minimizing waste.
The surface area function derived, \( S = xy + \frac{216}{x} + \frac{216}{y} \), captures well the amount of material used. We then focus on altering the dimensions \( x \), \( y \), and \( z \) to find the least value for this function. Minimizing surface area is pivotal, especially in manufacturing and packaging industries, where efficient use of materials can lead to significant economic and environmental benefits.

Partial Derivatives

Partial derivatives extend the concept of a derivative to multivariable functions. They measure how a function changes as one particular variable changes, while others are held constant. This concept is crucial in optimization problems with functions involving several variables.
In this problem, to minimize the surface area function \( f(x, y) \), we compute partial derivatives \( \frac{\partial f(x, y)}{\partial x} \) and \( \frac{\partial f(x, y)}{\partial y} \). These derivatives tell us how changes in \( x \) and \( y \) affect surface area. By setting these partial derivatives to zero, we locate critical points, which potentially identify the minimal surface area configuration for the box. This application supports multidimensional analysis, enabling optimization in more complex scenarios beyond single-variable calculus.

Volume Constraint

Volume constraint represents a fixed requirement that must be satisfied when designing an object. For this box, the constraint \( xyz = 108 \) ensures that every dimension combination results in the same volume. Constraints are common in real-world optimization problems, often reflecting finite resources or specific specifications.
In tackling such problems, constraints help shape potential solutions. By integrating the volume constraint into surface area calculations, we realize the dependency between dimensions like \( z \) as a function of \( x \) and \( y \). This linkage streamlines our approach, helping to eliminate variables and highlight feasible solutions.

• Ensuring the box maintains its volume while optimizing material use illustrates well-managed resources.
• Mathematical optimization under constraints simulates real-world conditions in engineering, architecture, and economics.

Understanding constraints enables solutions that effectively balance demands and limitations, a critical skill in practical applications.