Skip to content
Miscellaneous Exercise 7 · Q78

Q.Solve each of the following L.P.P. : Maximize z=2x+3yz = 2x + 3y subject to x−y≥3, x≥0, y≥0x - y \ge 3,\ x \ge 0,\ y \ge 0

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
78% · 78/100 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 →

The only finite corner point of x−y≥3, x≥0, y≥0x-y\ge3,\ x\ge0,\ y\ge0 is (3,0)(3,0) (where x−y=3x-y=3 meets y=0y=0). But moving along the boundary line x−y=3x-y=3 (i.e. x=y+3x=y+3) as y→∞y\to\infty stays feasible for every y≥0y\ge0 (since x=y+3≥0x=y+3\ge0 automatically), and z=2(y+3)+3y=5y+6→∞z=2(y+3)+3y=5y+6\to\infty as y→∞y\to\infty. So z=2x+3yz=2x+3y has no finite maximum on this feasible region — this is the same 'unbounded region, no optimal maximum' situation the chapter itself …

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.