1. **State the problem:** We need to find the shape of the largest rectangular beam that can be cut from a log of fixed size. The strength $S$ of the beam is proportional to the breadth $b$ and the square of the depth $d$, i.e., $$S \propto b d^2$$ and for the stiffest beam, strength (stiffness) is proportional to breadth and the cube of depth: $$S \propto b d^3$$ Given the log size restricts the amount of material, we assume a fixed perimeter or total dimension constraint. --- 2. **Define variables and constraint:** Let the total cross-sectional perimeter or boundary be fixed. For simplicity, assume the sum of breadth and depth is constant: $$b + d = c,$$ where $c$ is a positive constant determined by the log size. --- ### Part 1: Largest beam by strength proportional to $b d^2$ 3. **Express strength function:** $$S = k b d^2,$$ where $k$ is a positive constant, which we can ignore since it does not change the maximization. From the constraint, $b = c - d$, so $$S(d) = (c - d) d^2 = c d^2 - d^3.$$ --- 4. **Find critical points by differentiation:** $$\frac{dS}{dd} = 2 c d - 3 d^2.$$ Set derivative to zero: $$2 c d - 3 d^2 = 0 \implies d(2 c - 3 d) = 0.$$ Critical points are: $$d=0$$ or $$d=\frac{2 c}{3}.$$ --- 5. **Check which critical point gives maximum strength:** Second derivative: $$\frac{d^2 S}{d d^2} = 2 c - 6 d.$$ At $d=\frac{2 c}{3}$, $$\frac{d^2 S}{dd^2} = 2 c - 6 \times \frac{2 c}{3} = 2 c - 4 c = -2 c < 0,$$ so this is a maximum. --- 6. **Find corresponding breadth:** $$b = c - d = c - \frac{2 c}{3} = \frac{c}{3}.$$ So the largest beam shape ratio has breadth to depth: $$\frac{b}{d} = \frac{c/3}{2 c/3} = \frac{1}{2}.$$ --- ### Part 2: Stiffest beam with stiffness proportional to $b d^3$ 7. **Express stiffness function:** $$S = k b d^3,$$ Again, $b = c - d$, so $$S(d) = (c - d) d^3 = c d^3 - d^4.$$ --- 8. **Find critical points:** $$\frac{dS}{dd} = 3 c d^2 - 4 d^3 = d^2 (3 c - 4 d).$$ Set derivative to zero: $$d=0$$ or $$d=\frac{3 c}{4}.$$ --- 9. **Check second derivative at critical points:** $$\frac{d^2 S}{d d^2} = 6 c d - 12 d^2.$$ At $d=\frac{3 c}{4}$, $$6 c \times \frac{3 c}{4} - 12 \times \left(\frac{3 c}{4}\right)^2 = \frac{18 c^2}{4} - 12 \times \frac{9 c^2}{16} = 4.5 c^2 - 6.75 c^2 = -2.25 c^2 < 0,$$ so this is a maximum. --- 10. **Find breadth at max stiffness:** $$b = c - d = c - \frac{3 c}{4} = \frac{c}{4}.$$ Shape ratio for the stiffest beam: $$\frac{b}{d} = \frac{c/4}{3 c/4} = \frac{1}{3}.$$ --- **Final answer:** - For maximum strength proportional to $b d^2$, the ratio $$\boxed{b : d = 1 : 2}.$$ - For maximum stiffness proportional to $b d^3$, the ratio $$\boxed{b : d = 1 : 3}.$$