18 - 20 Rectangular beam in maxima and minima problems

Diameter is given (log of given size), thus D is constant
$b^2 + d^2 = D^2$

$2b\dfrac{db}{dd} + 2d = 0$

$\dfrac{db}{dd} = -\dfrac{d}{b}$
 

Strength:
$S = bd^2$

$\dfrac{dS}{dd} = b (2d) + d^2 \dfrac{db}{dd} = 0$

$2bd + d^2 \left( -\dfrac{d}{b} \right) = 0$

$2bd = \dfrac{d^3}{b}$

$2b^2 = d^2$

$d = \sqrt{2} \, b$
 

$\text{depth } \, = \sqrt{2} \, \times \, \text{ breadth}$           answer

Diameter is given (log of given size), thus D is constant
$b^2 + d^2 = D^2$

$2b\dfrac{db}{dd} + 2d = 0$

$\dfrac{db}{dd} = -\dfrac{d}{b}$
 

Stiffness:
$k = bd^3$

$\dfrac{dk}{dd} = b(3d^2) + d^3 \dfrac{db}{dd} = 0$

$3bd^2 + d^3 \left( -\dfrac{d}{b} \right) = 0$

$3bd^2 = \dfrac{d^4}{b}$

$3b^2 = d^2$

$d = \sqrt{3} \, b$
 

$\text{depth } \, = \sqrt{3} \, \times \, \text{ breadth}$           answer

Strength, S = bd²
Stiffness, k = bd³
 

For 2" × 8":

Oriented such that the breadth is 2"
S = 8(2²) = 32 in³
k = 8(2³) = 64 in⁴

Oriented such that the breadth is 8"
S = 2(8²) = 128 in³
k = 2(8³) = 1024 in⁴
 

For 4" × 6":

Oriented such that the breadth is 6"
S = 6(4²) = 96 in³
k = 6(4³) = 384 in⁴

Oriented such that the breadth is 4"
S = 4(6²) = 144 in³
k = 4(6³) = 864 in⁴
 

2" x 8" is stiffer than 4" x 6" and it is the commonly used size for floor joists. In fact, some local codes required a minimum depth of 8".