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)=−8x2+8x+30
- Z′(x)=−16x+8, hence Z′(x)=0 gives x=12. Z″(12)=−16 and hence, x=12 is a maximum location.
- y(12)=10−4⋅12=8 and z(12,8)=32.