80 Classical Optimization Techniques
Figure 2.7 Cross section of the log.This problem has two variables and one constraint; hence Eq. (2.25) can be applied
for finding the optimum solution. Since
f=kx−^1 y−^2 (E 1 )g=x^2 +y^2 −a^2 (E 2 )we have
∂f
∂x= −kx−^2 y−^2∂f
∂y= − 2 kx−^1 y−^3∂g
∂x= 2 x∂g
∂y= 2 yEquation (2.25) gives−kx−^2 y−^2 ( 2 y)+ 2 kx−^1 y−^3 ( 2 x)=0 at (x∗, y∗)thatis,y∗=√
2 x∗ (E 3 )Thus the beam of maximum tensile stress carrying capacity has a depth of√
2 times
its breadth. The optimum values ofxandycan be obtained from Eqs.(E 3 ) nda (E 2 )
as
x∗=a
√
3and y∗=√
2
a
√
3