MHT CET · Maths · Linear Programming
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
- A no solution.
- B a unique solution.
- C infinite solutions.
- D unbounded solutions.
Answer & Solution
Correct Answer
(C) infinite solutions.
Step-by-step Solution
Detailed explanation
| \(\text{line}\) | \(\text{Point on X- axis}\) | \(\text{Point on Y - axis}\) |
| \(x+y=30^ \circ\) | \(A(30,0)\) | \(B(0,30)\) |
| \(x=15\) | \(C(15,0)\) | \(-\) |
| \(y=20\) | \(-\) | \(D(0,20)\) |
| \(x+y=15\) | \(C(15,0)\) | \(F(0,15)\) |
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.
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