Problem 1 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Suppose you have a 3 meter by 4 meter sheet of tin and you wish to make a box that has tin on the bottom and on two opposite sides. The other two sides are of wood that is in plentiful supply. You are going to make a rectangular box by folding up panels of width \(x\) across the ends that are 3 meters wide as shown in Figure \(8.2 .1 .\) What value of \(x\) will maximize the volume of the box and what is the volume?

Step-by-Step Solution

Answer

The value of \(x\) that maximizes the volume is \(\frac{3}{4}\) meters, and the maximum volume is 4.5 cubic meters.

Step 1: Understand the Problem

You need to form a box with a base and two sides made of tin and the other two wooden sides. You will adjust the height, denoted by \(x\), to maximize volume.

Step 2: Write the Volume Expression

The base of the box is 3m by 4m. If you fold two sides of width \(x\), then the height is \(x\), the length is \(3-2x\), and the width remains 4m. Thus, the volume \(V\) is given by \(V = x \times 4 \times (3-2x) = 4x(3-2x)\).

Step 3: Simplify the Volume Function

Simplify \(V = 4x(3-2x)\) to get \(V = 12x - 8x^2\). This is the function to maximize.

Step 4: Find the Maximum Volume

Calculate the derivative of \(V\) with respect to \(x\): \(\frac{dV}{dx} = 12 - 16x\). Set this derivative to zero to find the critical points: \(12 - 16x = 0\), so \(x = \frac{12}{16} = \frac{3}{4}\) meters.

Step 5: Verify Maximization

To confirm that this critical point is a maximum, check the second derivative \(\frac{d^2V}{dx^2} = -16\). Since \(-16 < 0\), the function has a local maximum when \(x = \frac{3}{4}\).

Step 6: Calculate the Maximum Volume

Substitute \(x = \frac{3}{4}\) back into the volume function: \(V = 12(\frac{3}{4}) - 8(\frac{3}{4})^2 = 9 - 8 \times \frac{9}{16} = 9 - 4.5 = 4.5 \) cubic meters.

Key Concepts

Volume Maximization

To maximize the volume of a structure, you need to strategically choose dimensions that yield the largest possible space within. In this specific problem, the aim is to create a box with a mix of materials and maximize its volume through a specific dimension, denoted as \(x\). The volume maximization process stems from identifying proper balance between different geometric dimensions.

Remember, the goal is to adjust the box's height \(x\), so the whole box can contain as much as possible. This involves using the dimensions provided by the tin sheet—\(3\) meters by \(4\) meters—to determine the best configuration. By bending parts of the tin sheet, the task involves calculating the volume using definition \(V = \text{length} \times \text{width} \times \text{height}\). Thereby one understands that manipulating these measurements can achieve the desired maximum volume.

In practice:

• Calculate potential height \(x\), based on available material.
• Calculate the maximum effective volume the available measurements can produce.

Derivatives

Derivatives in calculus measure the rate at which a function changes. Here, they play a crucial role in optimization problems such as maximizing volume. Bringing derivatives into the scenario allows looking at how volume changes with minute alterations in dimension \(x\).

For volume maximization, the derivative of the volume function \(V\) with respect to \(x\) is essential. It expresses the rate at which volume changes by changing \(x\), important in optimization.

The process:

• Start with the volume function \(V = 12x - 8x^2\).
• Find its first derivative, \(\frac{dV}{dx} = 12 - 16x\).
• This derivative tells you how the volume reacts as dimensions adjust.

Critical Points

Critical points in calculus help identify where the graph or function's behavior changes. For volume maximization, critical points indicate potential maximums or minimums in the volume function.

You find these points by setting the derivative to zero, \(\frac{dV}{dx} = 0\). It shows where the slope of the volume function is zero—meaning the volume stops increasing or decreasing momentarily.

Key actions:

• Set \(\frac{dV}{dx} = 12 - 16x = 0\).
• Solve for \(x\) to identify critical value—the point to check further with the second derivative test.

Reach critical points to establish peaks in volume, necessary for determining if changes yield the best possible volume at \(x = \frac{3}{4}\) meters.

Second Derivative Test

Once a critical point is found, the second derivative test determines whether that point signifies a maximum or minimum. This involves the second derivative of the volume function, adding precision to outcomes.

In this task:

• The second derivative \(\frac{d^2V}{dx^2} = -16\) was calculated.
• Since it is less than zero \(-16 < 0\), the function has a local maximum where \(x = \frac{3}{4}\).

Why use this test?

• It confirms if the critical point is indeed where the maximum volume occurs.
• A negative result indicates a concave-downward shape—representative of a local maximum, crucial for ensuring the ultimate volume is reached.