Question

If we cut four congruent squares out of the corners of a square piece of cardboard 12 inches on a side, we can fold up the four remaining flaps to obtain a tray without a top. What size squares should be cut in order to maximize the volume of the tray? (See figure.)

Short answer

Answer: The optimal side length of the congruent squares to cut from each corner is 2 inches.

Step-by-step solution

Define variables and visualize the problem

Let x be the side length of the congruent squares being cut from each corner, and let V be the volume of the resulting tray. Our task is to maximize V. The tray will have a base with side lengths (12-2x) and a height of x, as shown on the figure.

Determine the tray volume

To calculate the volume of the tray, we can use the formula for the volume of a rectangular prism, which is V = length × width × height. In this case, the length and width are equal (12-2x), and the height is x. So the volume of the tray is: V(x) = (12-2x)(12-2x)x

Simplify the volume function

Now let's simplify the volume function V(x) by multiplying it out: V(x) = (144 - 48x + 4x^2)x = 4x^3 - 48x^2 + 144x.

Find the critical points

To find the critical points of the volume function, we must differentiate it with respect to x and set the result to zero: V'(x) = (12x^2 - 96x + 144) Now, we'll set V'(x) to zero and solve for x: 0 = 12x^2 - 96x + 144 Since the equation is quadratic, we can divide the equation by 12 to simplify it: 0 = x^2 - 8x + 12

Solve the quadratic equation

Now let's solve the quadratic equation x^2 - 8x + 12 = 0. We can factor it as (x - 2)(x - 6) = 0. Thus, we get our two solutions for x: x = 2 inches x = 6 inches

Determine which solution is relevant

Since we're cutting squares out of a 12-inch square piece of cardboard, an x value of 6 inches would exceed the maximum allowable size of congruent squares that can be cut from the corners. Therefore, we must choose x = 2 inches.

Conclusion

To maximize the volume of the tray, we should cut four congruent squares of side length 2 inches from each corner of the 12-inch square piece of cardboard.

Key concepts

Optimization in Microeconomics

Optimization is a fundamental concept in microeconomics, where individuals and firms are often modeled as making decisions that maximize their utility or profit. The idea is to find the best possible outcome given certain constraints, such as budget limitations or production capacities.

In the context of this exercise, the optimization problem involves finding the largest possible volume of a tray that can be created from a flat piece of cardboard by cutting out and folding up the sides. Here, the constraint is the size of the cardboard. Just like a firm deciding how much of a product to produce to maximize profit, we are deciding on the size of the squares to cut to maximize volume. By setting up a mathematical model, we can use techniques from calculus to solve for the optimal decision.

Applying Optimization to the Tray Problem

• We define our objective function: the volume of the tray, which we aim to maximize.
• Constraints come into play by the fixed size of the cardboard, limiting the possible size of the cut-out squares.
• The 'solution' is the size of the squares that yields the greatest volume when folded into a tray.

Similar optimization methods are used in microeconomics to make decisions about production levels, pricing, and resource allocation.

Quadratic Functions

Quadratic functions are polynomial functions of degree two and are typically written in the form f(x) = ax^2 + bx + c. They graph as parabolas which open upwards when a > 0 and downwards when a < 0. The vertex of the parabola represents either the maximum or minimum point, depending on the direction the parabola opens.

In the exercise, a quadratic function arises when we express the volume of the tray as a function of the size of the cut-out squares. When we simplify the expression for volume, we end up with a quadratic function in terms of x, the length of a side of the square, showing the relationship between x and the tray's volume.

Features of the Quadratic Function in this Problem

• The quadratic function is used to model the volume of the tray.
• Its graph will help us understand how changes in x affect the tray’s volume.
• We use the properties of quadratic functions to determine the maximum volume, which is the main goal of this exercise.

Understanding how to manipulate and analyze quadratic functions is essential in solving many applied mathematics problems, such as this one involving maximizing volume.

Calculus Derivatives

Derivatives are a cornerstone of calculus, representing the rate at which a function is changing at any given point. In practical terms, taking a derivative helps us understand how a small change in one variable affects another variable.

In optimization problems, like the one we're considering, derivatives play a critical role in finding maximum or minimum values of functions. To find these values, we calculate the derivative of the function with respect to the variable of interest and look for the points where this derivative is zero (these points are called 'critical points').

Using Derivatives to Solve the Problem

• We first differentiate the volume function with respect to x to find its derivative, V'(x).
• Setting V'(x) to zero allows us to find critical points that can be potential candidates for maximization.
• After finding the critical points, we evaluate them within the context of our problem to determine which one yields the maximum volume.

Derivatives make it possible to navigate complex optimization problems efficiently by pinpointing the exact moment when an increase turns into a decrease (or vice versa), thus identifying optimal values.