The stiffness of a beam of rectangular cross-section varies as the product of the breadth and square of the depth. What must be the breadth of the stiffest beam that can be cut from a leg of diameter 'd'?

Answer

To solve the problem, we need to find the breadth \( B \) of the stiffest beam that can be cut from a leg of diameter \( d \). The stiffness \( I \) of the beam is given to vary as the product of the breadth and the square of the depth.

### Step 1: Define the variables
Let:
- \( B \) = breadth of the beam
- \( H \) = depth of the beam
- \( d \) = diameter of the leg

### Step 2: Relate the breadth and depth to the diameter
Since the beam is rectangular and fits inside a circular leg of diameter \( d \), we can use the Pythagorean theorem:
\[ B^2 + H^2 = d^2 \]
From this, we can express \( H \) in terms of \( B \):
\[ H^2 = d^2 - B^2 \]

### Step 3: Express the stiffness
The stiffness \( I \) is proportional to the breadth and the square of the depth:
\[ I = k B H^2 \]
Substituting \( H^2 \) from the previous step:
\[ I = k B (d^2 - B^2) \]
This simplifies to:
\[ I = k (B d^2 - B^3) \]

### Step 4: Differentiate the stiffness with respect to breadth
To find the maximum stiffness, we differentiate \( I \) with respect to \( B \):
\[ \frac{dI}{dB} = k (d^2 - 3B^2) \]

### Step 5: Set the derivative to zero to find critical points
To find the critical points, set the derivative equal to zero:
\[ d^2 - 3B^2 = 0 \]
Solving for \( B \):
\[ 3B^2 = d^2 \quad \Rightarrow \quad B^2 = \frac{d^2}{3} \quad \Rightarrow \quad B = \frac{d}{\sqrt{3}} \]

### Step 6: Verify that this is a maximum
To confirm that this critical point gives a maximum, we can find the second derivative:
\[ \frac{d^2I}{dB^2} = -6kB \]
Since \( k > 0 \) and \( B > 0 \), we have:
\[ \frac{d^2I}{dB^2} < 0 \]
This indicates that the stiffness is maximized at \( B = \frac{d}{\sqrt{3}} \).

### Final Answer
The breadth of the stiffest beam that can be cut from a leg of diameter \( d \) is:
\[ B = \frac{d}{\sqrt{3}} \]