Problem 19 - Contemporary Precalculus

Question

A 20 -inch square piece of metal is to be used to make an open-top box by cutting equal-sized squares from each corner and folding up the sides (as in Example 3 on page 117). The length, width, and height of the box are each to be less than 14 inches. What size squares should be cut out to produce a box with (a) volume 550 cubic inches? (b) largest possible volume?

Step-by-Step Solution

Answer

Answer: To find the optimal size of the squares that should be cut from the corners, we first define x as the side length of the equal-sized squares. We then obtain the volume equation V(x) = x(20-2x)(20-2x) and analyze it within the constraints of x < 14 and 20-2x < 14. For a volume of 550 cubic inches, we set V(x) = 550 and solve for x. To find the largest possible volume, we identify the critical points of V(x) on the interval (0, 7) by solving V'(x) = 0 and analyze the function using the first and second derivative tests. Based on this analysis, we can determine the optimal size of the squares to be cut from the corners.

Step 1: Define variables and equations for volume and constraints

Let x be the side length of the equal-sized squares to be cut from each corner. The resulting dimensions of the box will be (20-2x) inches in length, (20-2x) inches in width, and x inches in height (due to folding up the sides). The volume of the box can be represented as V = x(20-2x)(20-2x). The given constraint is that the length, width, and height are each to be less than 14 inches, this can be expressed as: x < 14 and 20-2x < 14.

Step 2: Create an equation for volume as a function of x

Given the relation for volume, we have: V(x) = x(20-2x)(20-2x). We will use this equation to find the appropriate value of x for both parts of the question. (a) Volume is 550 cubic inches

Step 3: Solve the equation for x when V(x) = 550 cubic inches

Set V(x) = 550 and solve for x: 550 = x(20-2x)(20-2x). To find the value of x that meets the constraints, we can either solve this equation algebraically or graph V(x) and find the x-intercept corresponding to V(x) = 550. (b) Largest possible volume

Step 4: Find critical points of V(x) within the constraints

To find the largest possible volume, we need to identify the critical points of V(x) on the interval (0, 7) based on the constraints. We will find the critical points by taking the first derivative of the volume equation with respect to x and setting it equal to 0. V'(x) = \frac{d}{dx} [x(20-2x)(20-2x)]

Step 5: Solve for critical points and analyze the function

Now, solve for x in V'(x) = 0 to find the critical points. Then analyze the behavior of V(x) in the neighborhood of these critical points and at the endpoints of the interval (0, 7) using the first and second derivative tests.

Step 6: Determine the optimal value of x for largest volume

Based on the analysis in Step 5, we can determine the maximum value of x that meets the constraints. The size of the squares to be cut from the corners must be equal to this optimal x value.

Key Concepts

Volume of a Box

When creating an open-top box from a sheet of metal, we aim to find the best dimensions to achieve desired volumes. Here, a 20-inch square piece of metal is used, and by cutting equal-sized squares from each corner, we then fold the sides up to form the box. The key to finding the volume lies in understanding the dimensions. After cutting squares of side length \(x\) from each corner, the remaining pieces form the base with dimensions \((20-2x) \times (20-2x)\), and the height is \(x\). Thus, the volume \(V\) is given by the formula:
\[ V(x) = x(20-2x)(20-2x) \]
This equation captures the entire process of forming the box and calculating its volume.

Critical Points

Critical points are essential in optimization problems because they can indicate potential maximum or minimum values. To find the critical points of the box's volume function \(V(x)\), you first need to take its derivative. The derivative will help identify where the slope of the function is zero, which corresponds to possible peak volumes.
Calculating the derivative \(V'(x)\), you need to set it equal to zero and solve:

• Find where \( V'(x) = 0 \). These points may be local minima or maxima.
• Critical points need further analysis to determine if they represent a maximum or minimum value for volume.

Understanding critical points allows us to refine options for cuts to maximize the box's volume.

Derivative Test

In optimization, the derivative test helps ascertain the nature of critical points, like whether they are maxima or minima. Once the derivative \(V'(x)\) is calculated, you utilize the first and second derivative tests:

• First Derivative Test: Examines changes of \(V'(x)\) around critical points. A change from positive to negative signals a local maximum.
• Second Derivative Test: Involves \(V''(x)\). If \(V''(x) < 0\) at a critical point, it's a maximum, while \(V''(x) > 0\) indicates a minimum.

These tests are powerful tools for determining the optimal size of squares to cut, ensuring the resulting box has the largest possible volume.

Constraints in Optimization

Constraints are conditions that solutions must adhere to. For the metal box, the constraints arise from physical limitations on size. Specifically:

• The side of each cut square must be less than 14 inches: \(x < 14\).
• The resulting dimensions \((20-2x)\) must also be less than 14 inches to fit within the original sheet.

Together, these constraints define the feasible region for \(x\) where valid solutions can occur, typically identified as \(0 < x < 7\). While solving for maximum volume, these constraints ensure practicality, binding the solution to the problem's real-world context.