Problem 30 - Calculus for Biology and Medicine

Question

Find the maximum volume of a rectangular open (bottom and four sides, no top) box with surface area \(75 \mathrm{~m}^{2}\).

Step-by-Step Solution

Answer

Maximize the volume by setting up equations for surface area and volume, solve for variables, and use calculus to find the maximum.

Step 1: Define the Problem

We need to find the maximum volume of an open-topped rectangular box with a fixed surface area of \(75 \text{ m}^2\). This box will have a height \(h\), a length \(l\), and a width \(w\).

Step 2: Establish Equations for Surface Area and Volume

The surface area of the box is calculated by the formula: \(lw + 2lh + 2wh = 75\). The volume of the box is \(V = l \times w \times h\). Our goal is to express the volume \(V\) in terms of two variables, eventually reducing it to one variable for maximization.

Step 3: Solve for One Variable in Terms of Others

Using the surface area equation, solve for one variable. Let's solve for \(w\): \(w = \frac{75 - 2lh}{l + 2h}\). Substitute \(w\) back into the volume formula.

Step 4: Substitute Back into Volume Formula

Substitute \(w\) from the previous step into \(V = lwh\). The expression becomes \(V(l, h) = lh \left(\frac{75 - 2lh}{l + 2h}\right)\). Simplify the expression.

Step 5: Optimize Volume Function

Use calculus to find the maximum volume. Differentiate the volume \(V(l, h)\) with respect to one of the variables, say \(l\), and set the derivative to zero to solve for critical points. Repeat for \(h\) if necessary.

Step 6: Determine Maximum Volume

Calculate the critical points from the derivatives and test them, along with boundary values, to find which produces the maximum volume. Use the second derivative test to verify maximum volume conditions.

Key Concepts

Calculus Applications

Calculus serves as a powerful tool in optimization problems, which involve finding maximum or minimum values of a function. In our problem, the main goal is to maximize the volume of a rectangular box with a given surface area. We know the surface area is fixed, at 75 m², and we want to find the dimensions—length, width, and height—that maximize the box’s volume.

To accomplish this, we make use of derivatives, which help in identifying critical points. Critical points occur where the derivative of a function is zero or undefined, signaling potential maximum or minimum values within that domain. After setting up the volume function in terms of one variable, taking its derivative, and setting it to zero, we can solve for the variable at critical points.

Lastly, applying the second derivative test—checking whether the second derivative is positive or negative—will confirm whether these points yield a maximum or minimum. Negative values ensure a maximum volume.

Surface Area and Volume

Understanding the relationship between surface area and volume is crucial in solving this problem. When dealing with geometric figures—like our rectangular box—these properties are interconnected. In this exercise:

• Surface Area (SA): The measure of the total area that the surface of the object occupies. For our open-topped box, calculated as \( lw + 2lh + 2wh = 75 \).


• Volume (V): The measure of the space occupied within the object, expressed as \( V = l \times w \times h \). Our goal is to maximize this volume.

Solving this optimization problem involves converting the surface area constraint into a usable form for the volume function. Once we express the volume with fewer variables using the constraint, it can be optimized. Remember, the key connection here is that any change in the dimensions affects both surface area and volume, but the surface area remains fixed in this scenario.

Rectangular Box

The rectangular box is a common object of study in optimization problems due to its simple geometry, which can be broken down into length, width, and height. For this problem, we need to specifically consider how removing the top impacts the calculations. An open rectangular box means the top surface area isn't counted in the total surface area.

When solving problems involving a rectangular box:

• It's important to understand how each dimension plays a part in forming the box. The area equations and volume derive directly from these dimensions.


• Translate real-world constraints (like fixed surface area) into mathematical equations. This is pivotal in calculus and optimization, providing the foundation to explore maximum or minimum solutions effectively.

Every factor contributes to the final solution, and small changes in dimensions influence both surface area and volume values directly, making it crucial to follow a methodical approach in solving such optimization tasks.