Problem 93 - Precalculus: Functions and Graphs

Question

\(A\) cardboard box with an open top and a square bottom is to have a volume of \(25 \mathrm{ft}^{3}\). Use a table utility to determine the dimensions of the box to the nearest 0.1 foot that will minimize the amount of cardboard used to construct the box.

Step-by-Step Solution

Answer

The dimensions are approximately 3.7 ft by 3.7 ft by 1.8 ft.

Step 1: Define Variables and Formulas

Let's assume the side length of the square base is \( x \) feet and the height of the box is \( h \) feet. The volume \( V \) of the box is given by \( V = x^2h \). We know the volume is \( 25 \text{ ft}^3 \), so \( x^2h = 25 \). The surface area \( S \) of the box with an open top is \( S = x^2 + 4xh \). Our goal is to find \( x \) and \( h \) that minimize \( S \).

Step 2: Express Height in Terms of x

We need to express \( h \) in terms of \( x \) using the volume formula. From \( x^2h = 25 \), we find that \( h = \frac{25}{x^2} \). Now substitute this expression for \( h \) in the surface area formula.

Step 3: Substitute h into Surface Area Formula

Substitute \( h = \frac{25}{x^2} \) into the surface area formula: \[ S = x^2 + 4x \left(\frac{25}{x^2}\right) = x^2 + \frac{100}{x} \]. Now, we have \( S \) expressed solely in terms of \( x \).

Step 4: Find Critical Points of S

To minimize \( S \), take its derivative with respect to \( x \):\[ \frac{dS}{dx} = 2x - \frac{100}{x^2} \]. Set \( \frac{dS}{dx} = 0 \) to find critical points:\[ 2x - \frac{100}{x^2} = 0 \]Solving gives \( 2x^3 = 100 \), or \( x^3 = 50 \), so \( x = \sqrt[3]{50} \approx 3.684 \).

Step 5: Verify Minimum Condition

Use the second derivative test to verify whether this critical point is a minimum. Compute \( \frac{d^2S}{dx^2} \):\[ \frac{d^2S}{dx^2} = 2 + \frac{200}{x^3} \], evaluate at \( x = 3.684 \):Since \( \frac{d^2S}{dx^2} > 0 \) at \( x = 3.684 \), the surface area is minimized at this point.

Step 6: Calculate h and Verify

Calculate the height \( h \) using \( h = \frac{25}{x^2} \):\[ h = \frac{25}{(3.684)^2} \approx 1.846 \]. Check the volume: \( (3.684)^2 \times 1.846 \approx 25 \text{ ft}^3 \), confirming our dimensions are correct.

Key Concepts

Derivative Test

The derivative test is a fundamental technique in calculus used to find local maxima and minima of a function. Essentially, it helps us locate the lowest and highest points on a curve. Here, the main goal is to find the minimum surface area for the given volume of the box. We start with the surface area formula expressed in terms of one variable:

• Take the first derivative, \( \frac{dS}{dx} = 2x - \frac{100}{x^2} \), to find where the rate of change is zero.
• Set this derivative equal to zero to find possible critical points, which could be minima or maxima.

After finding the critical point \( x = \sqrt[3]{50} \), we apply the second derivative test to confirm its nature. Calculating the second derivative, \( \frac{d^2S}{dx^2} = 2 + \frac{200}{x^3} \), shows it is greater than zero at the critical point, indicating a local minimum. This test confirms that the surface area is indeed minimized at the computed value of \( x \).

Critical Points

Critical points are essential in the process of optimization as they mark where a function’s first derivative is zero or undefined. These points represent potential locations where the function could have a minimum or maximum value.

We found the critical points by setting the derivative of the surface area function to zero, resulting in the equation

• \( 2x - \frac{100}{x^2} = 0 \).
• Solving gives the critical point \( x = \sqrt[3]{50} \), which approximates to \( 3.684 \).

This critical point is examined further to determine the nature (minimum or maximum) using the second derivative, ensuring that the surface area is minimized efficiently.

Surface Area Minimization

Surface area minimization is an optimization problem where the goal is to find dimensions that use the least material, lowering costs and resources used. Given the open-top box, our aim is to minimize the cardboard needed.

Once \( h \) was expressed in terms of \( x \), we substituted it back into the surface area formula, simplifying the problem to one variable. Simplifying the function to \( S = x^2 + \frac{100}{x} \) helps focus directly on the dimensions influencing the surface area directly.

Through calculus techniques of taking derivatives and confirming via the second derivative test, the minimized configuration is determined, ensuring that the dimensions found indeed result in the least surface area required for the specified volume of the box.

Volume Constraints

Volume constraints play a significant role in design problems, as they define the limits within which the other parameters must work. In this problem, the box is required to have a volume of \( 25 \text{ ft}^3 \), balancing the need between dimension choices and resource usage.

• Volume is determined by the equation \( V = x^2h = 25 \).
• To link \( x \) and \( h \), we rearrange it as \( h = \frac{25}{x^2} \).

This relationship is crucial since it binds the dimensions and ensures the calculations for minimized surface area still satisfy the initial volume requirement. Properly applying and adhering to such constraints is essential in practical engineering and mathematical scenarios.