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
  1. y(x)=104x
  2. Z(x)=2x(104x)+3(104x)=8x2+8x+30
  3. Z(x)=16x+8, hence Z(x)=0 gives x=12. Z(12)=16 and hence, x=12 is a maximum location.
  4. y(12)=10412=8 and z(12,8)=32.
Conclusion: z(12,8)=32 is a maximum.