MinMax03 The strength of a rectangular wooden beam.

The strength of a rectangular wooden beam varies jointly as the width and cube of the depth of the beam. Find the dimensions of the strongest beam that can be cut from a log of radius R.

Sketch the round log and the rectangular beam. Do you remember word problems in algebra that contained phrases like "… varies jointly as … ?" This problem is included to remind you that some instructors use this language in calculus problems. The first statement in the problem, translated into algebra, is S = wd³. This is the defining equation.

To translate the problem statement completely, there should be a constant in front of the w but we are not going to calculate specific strengths, just the dimensions for maximum strength so the constant is not necessary. The constraint equation involves writing the Pythagorean statement for the right triangle formed by d, w, and 2R.

The constraint equation, d² + w² = 4R² can be solved for either d or w and substituted in the defining equation. Either way does not look too appealing. Solving for w keeps the numbers smaller so write w = (4R² - d²)^1/2 and substitute into the defining equation to write S in terms of d only.

S = (4R+ - d²)^1/2 d³

Differentiate S with the product rule.

S' = (4R² - d²)^1/2 (3d²) + d³(1/2)(4R² - d²)^-1/2 (-2d) = 3d² (4R² - d²)^1/2 - [d⁴ / (4R² - d²)^1/2]

Set S' equal to zero

3d² (4R² - d²)^1/2 = [d⁴ / (4R² - d²)^1/2] or d² = 3(4R² - d²) or 4d² = 12R²

and d = 3^1/2 R . The positive value for d substituted into the constraint equation produces w² = 4R² - 3R² = R² and w = R .

The maximum strength beam that can be cut from a log of radius R is one of dimensions R and 3^1/2 R . It is not necessary to formally determine that this is a maximum. It is the only reasonable choice from the first derivative equals zero condition.