MHT CET · Maths · Linear Programming
The maximum value of \(\mathrm{z}=4 x+2 y\), subject to the constraints \(3 x+4 y \geqslant 12, x+y \leqslant 5, x, y \geqslant 0\) is
- A 8
- B 20
- C 24
- D 16
Answer & Solution
Correct Answer
(B) 20
Step-by-step Solution
Detailed explanation
The feasible region. lies on the origin side of \(x+y=5\) and non-origin side of \(3 x+4 y=12\), in the first quadrant.
The corner points of the feasible region are \(\mathrm{A}(4,0), \mathrm{B}(5,0), \mathrm{C}(0,5)\) and \(\mathrm{D}(0,3)\).
\(z=4 x+2 y\)
At \(\mathrm{A}(4,0), \mathrm{z}=16\)
At B(5, 0), \(z=20\)
At C(0, 5), z=10
At \(D(0,3), z=6\)
\(\therefore \quad\) Maximum value of \(z\) is 20.
The corner points of the feasible region are \(\mathrm{A}(4,0), \mathrm{B}(5,0), \mathrm{C}(0,5)\) and \(\mathrm{D}(0,3)\).
\(z=4 x+2 y\)
At \(\mathrm{A}(4,0), \mathrm{z}=16\)
At B(5, 0), \(z=20\)
At C(0, 5), z=10
At \(D(0,3), z=6\)
\(\therefore \quad\) Maximum value of \(z\) is 20.
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