Skip to content
Question 90 of 100

Q.Minimize z=4x+5yz = 4x + 5y subject to 2x+y≥72x + y \ge 7, 2x+3y≤152x + 3y \le 15, x≤3x \le 3, x≥0x \ge 0, y≥0y \ge 0. Solve using graphical method.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2016Subjective· 4mImportance★★★★★
90% · 90/100 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 →

Plot the feasible region for the three inequalities, find its corner points, and evaluate zz at each.

Constraints: 2x+y≥72x+y\ge7, 2x+3y≤152x+3y\le15, x≤3x\le3, x≥0x\ge0, y≥0y\ge0.

Boundary lines:

  • L1:2x+y=7L_1: 2x+y=7
  • L2:2x+3y=15L_2: 2x+3y=15
  • L3:x=3L_3: x=3

Finding corner points of the feasible region (region satisfying 2x+y≥72x+y\ge7, i.e. above/right of L1L_1; 2x+3y≤152x+3y\le15, below L2L_2; and 0≤x≤30\le x\le3):

L1L_1 and L2L_2 intersect where: subtracting 2x+y=72x+y=7 from 2x+3y=152x+3y=15 gives 2y=8⇒y=42y=8\Rightarrow y=4, then 2x=7−4=3⇒x=1.52x=7-4=3\Rightarrow x=1.5. Point: (1.5, 4)(1.5,\,4).

L1L_1 and L3L_3 (x=3x=3): 2(3)+y=7⇒y=12(3)+y=7\Rightarrow y=1. Point: (3, 1)(3,\,1). [Check L2L_2: 2(3)+3(1)=9≤152(3)+3(1)=9\le15 ✓ feasible]

L2L_2 and L3L_3 (x=3x=3): 2(3)+3y=15⇒y=32(3)+3y=15\Rightarrow y=3. Point: (3, 3)(3,\,3). [Check L1L_1: 2(3)+3=9≥72(3)+3=9\ge7 ✓ feasible]

…

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.