The L.P.P. to maximize $z=x+y$, subject to $x+y \leq 30, x \leq 15, y \leq 20, x+y \geq 15$, $x, y \geq 0$ has

The L.P.P. to maximize $z=x+y$, subject to $x+y \leq 30, x \leq 15, y \leq 20, x+y \geq 15$, $x, y \geq 0$ has
  1. no solution.
  2. a unique solution.
  3. infinite solutions.
  4. unbounded solutions.

Solution

$\begin{array}{|l|l|l|} \hline \text{line} & \text{Point on X-axis} & \text{Point on y-axis} \\ \hline x+y=30^{\circ} & A(30,0) & B(0,30) \\ \hline x=15 & C(15,0) & - \\ \hline y=20 & - & D(0,20) \\ \hline x+y=15 & C(15,0) & F(0,15) \\ \hline \end{array}$ Point of intersection of $x=15$ and $y=20$ is $E \equiv(15,20)$ Feasible region is FCEDF. We have to maximize $Z=x+y$ $\begin{array}{l} Z_{(C)}=15+0=15 \\ Z_{(E)}=15+20=35 \\ Z_{(D)}=0+20=20 \\ Z_{(F)}=0+15=15 \end{array}$ Thus minimum value 15 occurs at two vertices $\mathrm{F}$ and $\mathrm{C}$. Thus given LPP has infinite solutions.

Asked in: MHT CET 2020 (13 Oct Shift 1)

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