We solve the following constrained extremum problem by the use of the substitution method.
- Maximize $z(x,y)=2xy+3y$
- Subject to $4x+y=10$
- Where $x,y>0$
- $y(x)=10-4x$
- $Z(x)=2x(10-4x)+3(10-4x)=-8x^2+8x+30$
- $Z'(x)=-16x+8$, hence $Z'(x)=0$ gives $x=\frac{1}{2}$. $Z''(\frac{1}{2})=-16$ and hence, $x=\frac{1}{2}$ is a maximum location.
- $y(\frac{1}{2})=10-4\cdot \frac{1}{2}=8$ and $z(\frac{1}{2},8)=32$.