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

Answer: (C)

$(4,\,\,4)$

Answer 2

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{aligned} & 4x+2y=24 \\\ & 3x+2y=20 \\\ & -\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,- \\\ & \\_\\_\\_\\_\\_\\_\\_\\_\\_\\_ \\\ & -5x=-6 \\\ \end{aligned} $\Rightarrow$ $x=\frac{6}{5}$ 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.