Problem 18 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

A rectangular wood beam with one side vertical has a 'stiffness' that is proportional to its width and the cube of its depth. A sawyer is to cut a single beam from a 1 meter diameter log. What dimensions should he cut the beam in order to maximize the stiffness of the beam?

Step-by-Step Solution

Answer

Cut the beam with width and depth as \( \frac{\sqrt{2}}{2} \) meters for maximum stiffness.

Step 1: Understand the Problem

We are asked to find the dimensions of a rectangular beam to maximize its stiffness. The stiffness, S, is proportional to the width, W, and the cube of the depth, D. Therefore, the stiffness can be represented as: \( S = k imes W imes D^3 \) where \( k \) is a constant of proportionality.

Step 2: Formulate the Constraint

The beam is to be cut from a 1-meter diameter log, which is a cylinder. The diagonal of the rectangle (beam) can be at most 1 m as it fits within the circle of diameter 1 m. Hence, for a rectangle with sides W and D, we have: \( W^2 + D^2 = 1 \).

Step 3: Express Objective Function

The objective is to maximize stiffness, which is \( S = k \times W imes D^3 \). We need to express \( W \) in terms of \( D \) (or vice-versa) using the constraint: \( W = \sqrt{1 - D^2} \). Thus, \( S = k \times \sqrt{1 - D^2} \times D^3 \).

Step 4: Differentiate and Find Critical Points

Differentiate \( S \) with respect to \( D \) and set the derivative equal to zero to find critical points. Compute: \( \frac{dS}{dD} = k \times \left( \frac{3D^2\sqrt{1-D^2} - D^4}{\sqrt{1-D^2}} \right) = 0 \). Simplifying gives \( 3D^2 = D^2(1-D^2) \), hence \( 2D^2 = D^4 \).

Step 5: Solve for D and Calculate W

Solving \( 2D^2 = D^4 \), divide by \( D^2 \) (assuming \( D eq 0 \)), provides \( D^2 = 2 \), thus \( D = \frac{\sqrt{2}}{2} \). Calculate \( W = \sqrt{1 - D^2} = \sqrt{1 - \left(\frac{\sqrt{2}}{2}\right)^2} = \frac{\sqrt{2}}{2} \).

Step 6: Confirm with Second Derivative Test

Use the second derivative to ensure the critical point is a maximum. The second derivative, \( \frac{d^2S}{dD^2} \), should be negative at \( D = \frac{\sqrt{2}}{2} \), confirming a maximum.

Key Concepts

Rectangular Beam

When dealing with a rectangular beam, imagine a long piece of wood with a rectangular cross-section. The beam's dimensions, namely its width and depth, are crucial because they directly influence the beam's functional properties, like stiffness, which is the ability to resist bending under load. In this context, we look at a beam that fits within the circular cross-section of a log with a 1-meter diameter. Because the beam is cut from this log, the diagonal of its rectangle cannot exceed the diameter of the circle. Thus, the maximum length for the diagonal is 1 meter. This geometric limitation provides us with a crucial constraint, expressed through Pythagoras's theorem: the sum of the squares of the width and depth equals the square of the diagonal. This essential constraint ensures that the beam can indeed fit within the circumference of the log.

Stiffness Maximization

Stiffness is a key property of structural elements like beams, describing how much they resist deformation under stress. For a rectangular beam, stiffness can be optimized by adjusting its width and depth. In our exercise, stiffness, denoted as \(S\), is proportional to both the width \(W\) and the cube of the depth \(D\).

• Width affects stiffness linearly: Doubling the width will double the stiffness.
• Depth has a cubic effect: Doubling the depth will increase stiffness eightfold.

This means depth is more influential in determining stiffness. However, given the constraint from the log's diameter, we must balance these dimensions. Our task is thus to find the optimal dimensions, using calculus, to achieve maximum stiffness while respecting the geometric limits imposed by the log's size.

Calculus Problem Solving

Calculus allows us to handle problems involving dynamic changes and optimization, such as finding dimensions that maximize beam stiffness. Here, calculus is used to maximize \(S = k \times W \times D^3\), subject to the constraint \(W^2 + D^2 = 1\). By substituting \(W = \sqrt{1-D^2}\) into the stiffness equation, we reduce the problem to one variable, making it manageable. Next, we calculate the derivative of \(S\) with respect to \(D\) to find the critical points. These are points where the rate of change, or the slope, is zero, indicating potential maxima or minima. By solving the derivative equation, we determine that for stiffness:

• \(D^2 = \frac{1}{2}\)
• \(W = \sqrt{1-D^2}\)

The second derivative test further assists in confirming we have found a maximum. Consequently, calculus helps us achieve the most efficient use of material by indicating the dimensions that give maximum stiffness, ensuring the beam stands up to forces without bending excessively.