question_answer

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)  \[(0,\,\,10)\]

Correct Answer: C

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.