Problem 12 - Essential Calculus Early Transcendentals

Question

A box with a square base and open top must have a volume of \(32,000 \mathrm{cm}^{3} .\) Find the dimensions of the box that minimize the amount of material used.

Step-by-Step Solution

Answer

Base: 40 cm, Height: 20 cm.

Step 1: Define Variables

Let the side of the square base be \( x \) cm and the height of the box be \( h \) cm. The volume of the box is given by the formula:\[V = x^2 h = 32,000\] cm³. Our task is to find the dimensions that minimize the surface area since we want to minimize the material used.

Step 2: Express Height in Terms of Base Side

Using the given volume constraint, express \( h \) in terms of \( x \):\[h = \frac{32,000}{x^2}\].

Step 3: Write Surface Area Formula

The surface area \( S \) for the box (with an open top) is the area of the four sides plus the base:\[S = x^2 + 4xh\]Substitute for \( h \) from Step 2:\[S = x^2 + 4x\left(\frac{32,000}{x^2}\right)= x^2 + \frac{128,000}{x}\]

Step 4: Optimize Surface Area

To minimize the surface area, find the derivative of \( S \) with respect to \( x \) and set it to zero to find critical points:\[\frac{dS}{dx} = 2x - \frac{128,000}{x^2}\]Set \( \frac{dS}{dx} = 0 \):\[2x - \frac{128,000}{x^2} = 0\]Solve for \( x \):\[2x^3 = 128,000\]\[x^3 = 64,000\]\[x = 40\] cm.

Step 5: Calculate Height

Substitute \( x = 40 \) cm back into the equation for \( h \):\[h = \frac{32,000}{40^2}= \frac{32,000}{1600}= 20\] cm.

Step 6: Verify Minimal Surface Area

To confirm that this indeed yields a minimum, check the second derivative:\[\frac{d^2S}{dx^2} = 2 + \frac{256,000}{x^3}\]When \( x = 40 \), \( \frac{d^2S}{dx^2} > 0 \), indicating a minimum.

Step 7: Conclusion: Dimensions of Box

The side of the square base is 40 cm, and the height of the box is 20 cm.

Key Concepts

Surface Area Minimization

When optimizing a box's surface area, the goal is to use the least amount of material, which is crucial in manufacturing to reduce costs. We need to look at the total area of the box's surfaces, which includes the base and sides, since the box has an open top. This can be expressed mathematically as the sum of these areas. With a defined volume, the challenge is to express the surface area purely in terms of one variable.

The equation for surface area, in this case, becomes:

• Area of base: \( x^2 \)
• Area of sides: \( 4xh \)

By substituting \( h \) from the volume equation, the surface area reflects both base and height, leading to a simplified expression that allows for easier optimization. Once the surface area is expressed in this manner, calculus provides a way to find the optimal dimensions that minimize the material.

Derivative Applications

Derivatives are a foundational tool in calculus, allowing us to find rates of change and identify extreme values. In optimization problems like minimizing the surface area of a box, we use derivatives to determine where a function increases or decreases.

First, we need to find the derivative of the surface area function \( S \) with respect to \( x \). This involves differentiating each term in the surface area equation:

• The derivative of \( x^2 \) is \( 2x \).
• The derivative of \( \frac{128,000}{x} \) is \( -\frac{128,000}{x^2} \).

Setting this derivative equal to zero, \( \frac{dS}{dx} = 0 \), helps locate potential minimum points by solving for \( x \). This is because a derivative equal to zero indicates a flat point where the function might present a minimum or maximum.

Critical Points Analysis

Identifying critical points is essential in determining the spots where our function behaves minimally or maximally. For surface area minimization, after finding the first derivative and setting it to zero, the solutions represent potential critical points.

In our example, solving \( 2x - \frac{128,000}{x^2} = 0 \) yields \( x = 40 \) cm. This is a candidate where the function might reach a minimum. However, merely finding critical points isn’t enough; we need to confirm whether these points are indeed minimums or maximums.

This confirmation can be performed using the second derivative test. A positive second derivative \( \frac{d^2S}{dx^2} > 0 \) indicates a point of minimum surface area. This test assures that choosing \( x = 40 \) cm leads to the least material used, fulfilling the optimization problem's requirements.