Best rectangle from a circular log – section for maximum bending strength From a circular wooden log of diameter 100 cm, a rectangular beam is cut. What breadth b and depth d (in cm) maximize the bending strength about the strong axis?

Introduction / Context:
When cutting a rectangular section from a given circular log, the goal is often to maximize bending strength (section modulus) about the strong axis. This is a classical optimization problem with a geometric constraint.

Given Data / Assumptions:

• Circular log diameter D = 100 cm.
• Rectangle breadth b and depth d inscribed in the circle → b^2 + d^2 = D^2.
• Strength ∝ section modulus about strong axis, Z = b d^2 / 6 for a rectangle.

Concept / Approach:
Maximize Z = b d^2 / 6 subject to the constraint b^2 + d^2 = D^2. Use calculus (substitute b = √(D^2 − d^2), differentiate Z with respect to d, set derivative to zero) to find the optimal ratio.

Step-by-Step Solution:
Constraint: b = √(D^2 − d^2).Z(d) ∝ d^2 √(D^2 − d^2).dZ/dd = 0 → 2D^2 − 3d^2 = 0 → d = (√6 / 3) D ≈ 0.8165 D.Then b = √(D^2 − d^2) = (√3 / 3) D ≈ 0.5774 D.For D = 100 cm: d ≈ 81.65 cm and b ≈ 57.73 cm.

Verification / Alternative check:
The ratio b : d = √3 : √6 = 1 : √2, a well-known result for the strongest rectangle from a circle.

Why Other Options Are Wrong:

• (b) swaps the axes; yields smaller Z about the strong axis.
• (c), (d), (e) do not satisfy the optimal ratio and give lower section modulus.

Common Pitfalls:
Maximizing area instead of section modulus; ignoring the circle-inscription constraint b^2 + d^2 = D^2.

Final Answer:
b = 57.73 cm, d = 81.65 cm