MHT CET · Maths · Linear Programming
The maximum value of \(z=7 x+8 y\) subject to the constraints \(x+y \leq 20, y \geq 5, x \leq 10, x \geq 0, y \geq 0\) is
- A \(150\)
- B \(160\)
- C \(110\)
- D \(180\)
Answer & Solution
Correct Answer
(B) \(160\)
Step-by-step Solution
Detailed explanation
Feasible region lies on the origin side of lines \(x+y=20, x=10\) and on non-origin side of \(y=5\).
\(\therefore \quad\) Corner points of the feasible region are \(\mathrm{A}(0,5), \mathrm{B}(10,5), \mathrm{C}(10,10)\) and \(\mathrm{D}(0,20)\)
\(\mathrm{z}\) at \(\mathrm{A}(0,5)=40\)
\(\mathrm{z}\) at \(\mathrm{B}(10,5)=110\)
\(\mathrm{z}\) at \(\mathrm{C}(10,10)=150\)
\(\mathrm{z}\) at \(\mathrm{D}(0,20)=160\)
\(\therefore \quad\) Maximum value of \(\mathrm{z}\) is 160 .
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