Skip to content
Miscellaneous Exercise 7 · Q81

Q.Solve each of the following L.P.P. : Maximize z=4x+2yz = 4x + 2y subject to 3x+y≥27, x+y≥21, x+2y≥30; x>0, y>03x + y \ge 27,\ x + y \ge 21,\ x + 2y \ge 30;\ x > 0,\ y > 0

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
81% · 81/100 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

This repeats part 5)ii) with one extra constraint, x+2y≥30x+2y\ge30, added. The feasible region of 3x+y≥27, x+y≥21, x+2y≥30, x>0, y>03x+y\ge27,\ x+y\ge21,\ x+2y\ge30,\ x>0,\ y>0 has corner points (30,0)(30,0), (12,9)(12,9), (3,18)(3,18), (0,27)(0,27), and remains unbounded upward for the same reason as before (three lower-bound-only constraints, no upper cap). Taking (x,y)=(100,0)(x,y)=(100,0) still satisfies all three constraints and gives z=400z=400, growing without bound — so as a literal maximization this has no finite answer. Reading it as Minimize z=4x+2yz=4x+2y instead: z(30,0)=120z(30,0)=120, z(12,9)=48+18=66z(12,9)=48+18=66, z(3,18)=12+36=48z(3,18)=12+36=48, z(0,27)=54z(0,27)=54 — the minimum is still z=48z=48 at (3,18)(3,18) (the extra constraint x+2y≥30x+2y\ge30 is satisfied the …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.