MHT CET · Maths · Linear Programming
The maximum value of \(\mathrm{z}=x+y\), subjected to \(x+y \leq 10,5 x+3 y \geq 15, x \leq 6, x, y \geq 0\)
- A occurs only at unique point.
- B occurs only at two distinct points.
- C occurs at infinitely many points.
- D does not exist.
Answer & Solution
Correct Answer
(C) occurs at infinitely many points.
Step-by-step Solution
Detailed explanation
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 \((0,10), \mathrm{C}(6,4), \dot{D}(6,0), \mathrm{E}(3,0)\)
At A(0,5), z = 0+5=5
At \(B(0,10), z=0+10=10\)
At C(6,4), \(z=6+4=10\)
At \(\mathrm{D}(6,0), \mathrm{z}=6+0=6\)
At \(\mathrm{E}(3,0), \mathrm{z}=3+0=3\)
\(\therefore \quad z\) has maximum value at \(\mathrm{B}(0,10)\) and \(C(6,4)\).
\(\therefore \quad z\) has infinite solution on seg BC.
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