Problem 13 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

A rectangular box with square base and with top and bottom and a shelf entirely across the interior is to have \(12 \mathrm{~m}^{3}\) volume. What dimensions of the box will minimize the material used?

Step-by-Step Solution

Answer

The box has dimensions 2 m by 2 m by 3 m.

Step 1: Define the Variables

Let's define the dimensions of the box. Let the side length of the square base be \( x \) (in meters), and the height of the box be \( h \). We know the box has a volume of \( 12 \ \mathrm{m}^3 \). This gives us the equation \( x^2 \, h = 12 \). The shelf divides the height \( h \) into two equal parts, so each part is \( \frac{h}{2} \).

Step 2: Express Height in Terms of the Side Length

From the volume equation \( x^2 \, h = 12 \), we can solve for height \( h \) in terms of \( x \): \( h = \frac{12}{x^2} \).

Step 3: Express the Material Used

The material used will be the surface area of the box, including the shelf. The box has a base and top (each \( x^2 \)), four vertical faces (each \( x \times h \)), and the shelf (\( x^2 \)). Thus the total surface area \( S \) is \( S = 2x^2 + 4xh + x^2 = 3x^2 + 4x \frac{12}{x^2} \).

Step 4: Simplify the Surface Area Expression

Simplify \( S = 3x^2 + \frac{48}{x} \). Here, \( 4x \frac{12}{x^2} = \frac{48}{x} \).

Step 5: Find the Critical Points

To minimize \( S \), find \( \frac{dS}{dx} \) and set it to zero. First, compute \( \frac{dS}{dx} \): \( \frac{dS}{dx} = 6x - \frac{48}{x^2} \). Set \( 6x - \frac{48}{x^2} = 0 \) and solve for \( x \).

Step 6: Solve for x

Solve \( 6x^3 = 48 \) which simplifies to \( x^3 = 8 \), giving \( x = 2 \).

Step 7: Determine Height h

Use the expression \( h = \frac{12}{x^2} \) with \( x = 2 \). Then \( h = \frac{12}{4} = 3 \).

Step 8: Check for Minimization

Verify if \( x = 2 \) minimizes \( S \) by checking the second derivative \( \frac{d^2S}{dx^2} \). Calculate \( \frac{d^2S}{dx^2} = 6 + \frac{96}{x^3} \). Since \( \frac{d^2S}{dx^2} > 0 \) at \( x = 2 \), it confirms a minimum.

Key Concepts

Volume constraints

In the context of optimizing a rectangular box with a square base, a key aspect to consider is the volume constraint. This constraint ensures that the box's volume remains constant throughout the process. Given in the exercise is a box that must maintain a specific volume of \( 12 \ \mathrm{m}^3 \). The volume of such a box can be defined by the equation \( x^2 \times h = 12 \), where \( x \) represents the side length of the square base, and \( h \) is the height of the box.
Understanding this constraint is crucial because it relates the box’s dimensions. You must express one dimension in terms of another using this volume equation. Here, solving for \( h \) in terms of \( x \) gives \( h = \frac{12}{x^2} \).

• This method narrows down the variables when optimizing other properties like surface area.
• It ensures the structural requirement that the volume should always equal \( 12 \ \mathrm{m}^3 \) is satisfied.

By keeping the volume constant, we can focus on adjusting dimensions to optimize other criteria, such as minimizing material use for constructing the box.

Surface area

Once volume constraints are in place, the next step is to minimize the surface area of the box, which directly relates to minimizing the material used in its construction. The surface area has components including the base, top, sides, and the shelf within the box.
The surface area \( S \) is given by \[ S = 2x^2 + 4xh + x^2 \], representing:

• Two squares for the base and top \(2x^2\).
• Four rectangular sides \(4xh\).
• The shelf inside, also a square \(x^2\).

By substituting \( h = \frac{12}{x^2} \) from the volume constraint equation, the surface area becomes \( S = 3x^2 + \frac{48}{x} \).
Simplifying the surface area equation allows focusing solely on \( x \), the side length of the base, making it manageable to find the minimum material required. Effective calculation and simplification of the surface area expression are critical in solving optimization problems efficiently.

Critical points

In calculus, finding critical points is essential for determining where a function reaches its minimum or maximum values. Critical points occur where the derivative is zero (indicating a horizontal tangent line) or undefined.
In our box optimization problem, we find these points by taking the derivative of the surface area expression \( S = 3x^2 + \frac{48}{x} \). The derivative \( \frac{dS}{dx} \) is calculated as \( 6x - \frac{48}{x^2} \). We set \( \frac{dS}{dx} = 0 \) to find where the function may have a minimum.

• Solving \( 6x - \frac{48}{x^2} = 0 \) simplifies to \( x^3 = 8 \), leading to \( x = 2 \).
• This indicates a potential point where the surface area is minimized.

To verify that this critical point provides a minimum surface area, we check the second derivative \( \frac{d^2S}{dx^2} \), calculated as \( 6 + \frac{96}{x^3} \). Since this second derivative is positive at \( x = 2 \), it confirms a local minimum.
Understanding how to identify and verify critical points is a crucial skill in optimization, allowing for efficient and accurate results.