Question:

Which of the following is/are solution of the LPP? \[ \text{Min }Z=12x+9y \] Subject to \[ x+2y\leq 40,\quad 3x+y\geq 30,\quad 4x+3y\geq 60,\quad x,y\geq 0 \] A. \((15,0)\),
B. \((40,0)\),
C. \((4,18)\),
D. \((6,12)\),
E. \((10.5,6)\).

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For LPP optimum questions, substitute each given point in the objective function and compare the values.
Updated On: Jun 6, 2026
  • A only
  • A and D only
  • A, D and E only
  • B and C only
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The Correct Option is C

Solution and Explanation

Concept:
In a Linear Programming Problem, the optimum value is checked at feasible corner points. Here, we calculate \(Z=12x+9y\) at the given feasible points.

Step 1: Calculate \(Z\) at point A \((15,0)\).
\[ Z=12(15)+9(0)=180 \]

Step 2: Calculate \(Z\) at point B \((40,0)\).
\[ Z=12(40)+9(0)=480 \]

Step 3: Calculate \(Z\) at point C \((4,18)\).
\[ Z=12(4)+9(18) \] \[ Z=48+162=210 \]

Step 4: Calculate \(Z\) at point D \((6,12)\).
\[ Z=12(6)+9(12) \] \[ Z=72+108=180 \]

Step 5: Calculate \(Z\) at point E \((10.5,6)\).
\[ Z=12(10.5)+9(6) \] \[ Z=126+54=180 \]

Step 6: Compare values.

Minimum value is: \[ Z_{\min}=180 \] This occurs at: \[ A,\ D,\ E \] \[ \therefore \text{Correct Answer is (C)} \]
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