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  4. 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

Q. 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

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Solution:

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.
Equations
Points
$ 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 method
Points
Objective 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.

Solution Image

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