The maximum value of $\mathrm{Z}=x+y$, subjected to $x+y \leq 10,5 x+3 y \geq 15, x \leq 6, x, y \geq 0$
- occurs only at unique point
- occurs only at two distinct points
- occurs at infinitely many points
- does not exist
Solution
Feasible region lies on the origin side of $x+y=10, x=6$ and non-origin side of $5 x+3 y=15$ The corner points of feasible region are $\mathrm{A}(0,5)$ and $\mathrm{B}(0,10), \mathrm{C}(6,4), \mathrm{D}(6,0), \mathrm{E}(3,0)$ At $\mathrm{A}(0,5), \mathrm{z}=0+5=5$ At B( 0,10$), \mathrm{z}=0+10=10$ At C(6,4), $z=6+4=10$ At $\mathrm{D}(6,0), \mathrm{z}=6+0=6$ At $E(3,0), z=3+0=3$ $\therefore \quad \mathrm{z}$ has maximum value at $\mathrm{B}(0,10)$ and $\mathrm{C}(6,4)$. $\therefore \quad \mathrm{z}$ has infinite solution on seg BC .
Asked in: MHT CET 2024 (03 May Shift 2)