Question

For the given LPP (Linear Programming Problem) max $ z=5x+3y $ $ 2x+y\le 12 $ $ 3x+2y\le 20 $ $ x\ge 0,\,y\ge 0 $ the optimal solution set is

• A. $ (0,\,\,0) $
• B. $ (6,\,\,0) $
• C. $ (4,\,\,4) $
• D. $ (0,\,\,10) $

Answer 1

The Correct Option is C

Given LPP is $ Max\,z=5x+3y $ $ 2x+y\le 12 $ $ 3x+2y\le 20 $ $ x\ge 0,\,\,\,\,\,\,\,\,\,y\ge 0 $ First we consider all the inequalities as equations.EquationsPoints $ 2x+y=12 $ $ (0,\,\,12),\,\,(6,0) $ and $ (0,\,\,10),\,\,\,\,\left( $ 3x+2y=20 $ \frac{20}{3},0 \right) $ Now, plot all these points on a graph paper and make a figure.For intersection point P, solve both equation of lines, we get $ \begin{align} & 4x+2y=24 \\ & 3x+2y=20 \\ & -\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,- \\ & \_\_\_\_\_\_\_\_\_\_ \\ & -5x=-6 \\ \end{align} $ $ \Rightarrow $ $ x=\frac{6}{5} $ and then $ y=\frac{22}{5} $ $ \therefore $ Convex region is TSQD with extreme point $ T(0,8),\,\,S(1,5),\,Q(2,4) $ and $ D(10,0) $ . Now, apply corner point methodPointsObjective function Max $ Z=5x+3y $ $ O(0,0) $ $ 5\times 0+3\times 0=0 $ $ B(0,10) $ $ 5\times 0+3\times 10=30 $ $ P(4,4) $ $ 5\times 4+3\times 4=32(\max ) $ $ A(6,0) $ $ 5\times 6+3\times 0=30 $ $ \therefore $ Optimal solution set is $ (4,4) $ on which the objective function is maximize.