Question

An open box of maximum volume is to be made from a square piece of material 24 centimeters on a side by cutting equal squares from the corners and turning up the sides.

Short answer

The maximum volume of the box that can be formed from the given material is 2560 cm\(^3\). This happens when the squares cut from the corners are of side length 4 cm.

Step-by-step solution

Understand the Setup

First, visualize the problem. You have a square piece of material of side length 24 cm. Squares will be cut out from each corner and the sides will be folded up to form a box. Let the side length of the squares cut out be \( x \) cm. Therefore, the length, width, and height of the box will be \( (24-2x) \), \( (24-2x) \), and \( x \) cm respectively.

Write Down the Volume Equation

The volume, \( V \), of a box can be calculated by multiplying its length, width and height. So, in this case it would be \( V = x(24-2x)^2 \).

Differentiate the Volume Equation and Find the Critical Numbers

To find maximum volume, differentiate the volume equation with respect to \( x \) and set it equal to zero, that is, \( V' = 0 \). Doing so yields \( V'(x) = 3*(24-2x)^2 - 4*x*(24-2x) \). Solving this for \( x \) using standard methods of solving equations will give us the critical numbers. In this case, \( x = 4 \) and \( x = 6 \) are obtained.

Determine the Maximum Value

Use the second derivative test to determine the maximum value. The second derivative of \( V \), \( V''(x) = -24(24-2x)+8x \). Evaluating this at \( x = 4 \) and \( x = 6 \), and taking into consideration that \( x \) cannot exceed the value \( 12 \) (due to the size limit of the original material), we see that the maximum volume is reached when \( x = 4 \) cm.

Calculate Maximum Volume

Substitute \( x = 4 \) back into the volume equation \( V = x(24-2x)^2 \) to obtain the maximum volume. This yields \( V = 4(24 - 2*4)^2 = 2560 \) cm\(^3\).

Key concepts

Optimization in Calculus

Optimization is a fundamental application of calculus, which involves finding the 'best', 'highest', or 'lowest' values that a function can achieve, often under certain constraints. In terms of geometry and real-world problems, this concept can be used to maximize or minimize quantities such as area, volume, cost, and efficiency. The optimization process typically involves defining a function that models the problem, finding its critical points by taking the first derivative and setting it to zero, and then determining whether these critical points are maxima or minima using methods such as the second derivative test. The open box problem is a classic example of optimization, where we aim to maximize the volume of the box formed from a fixed-size piece of material.

Volume Calculation

Volume calculation is at the heart of the open box problem. It's essential to express the volume of the object you're creating or studying as a function of its varying dimensions. In this case, the box's volume is modeled by the function \( V = x(24-2x)^2 \), which comes from the formula for the volume of a rectangular prism (Volume = length \'97width \'97height). The dimensions of the box change when corners are cut out and the sides are folded up, so we use 'x' to represent the size of the squares cut from each corner. This function captures the essence of the problem and sets us up to find the optimal 'x' that maximizes the volume.

Differentiation Applications

Differentiation is a powerful tool in calculus used to find how a function changes at any given point. When it comes to optimization problems, differentiation helps identify the critical points where the function achieves a maximum or minimum value. By differentiating the volume function relative to 'x', we obtain \( V'(x) = 3*(24-2x)^2 - 4*x*(24-2x) \), which can be set to zero to find the values of 'x' that might lead to the maximum volume. Application of differentiation is not only crucial for finding where the maximum volume occurs but also for understanding the behavior of the function across different values of 'x'.

Critical Numbers in Calculus

Critical numbers are points at which a function's derivative is zero or undefined. They are potential candidates for local maxima or minima of the function. In the context of our open box problem, after differentiating the volume function, we find critical numbers by solving for 'x' in the equation \( V'(x) = 0 \). We found that the values \( x = 4 \) and \( x = 6 \) are the critical numbers for this particular function. However, not all critical numbers will yield the maximum value we seek; it is essential to test these critical points further to determine which one truly maximizes the volume.

Second Derivative Test

The second derivative test is a method used to classify the critical points of a function as local maxima, minima, or points of inflection. This test involves computing the second derivative of the function and evaluating it at the critical numbers. A positive value indicates the function has a local minimum at that point, and a negative value suggests a local maximum. For the open box problem, applying the second derivative test to the critical numbers \( x = 4 \) and \( x = 6 \) helps us to determine that \( x = 4 \) indeed maximizes the volume, as the second derivative is negative, which signifies a concave down curve at this point, indicative of a maximum value. This step is key to confirming that our solution will give us the largest possible box volume within the constraints of the original piece of material.