MHT CET · Maths · Linear Programming
Maximum value of \(\mathrm{Z}=100 x+70 y\) Subject to \(2 x \geq 4, y \leq 3, x+y \leq 8, x, y \geq 0\) is
- A 800 .
- B 940 .
- C 400 .
- D 710 .
Answer & Solution
Correct Answer
(A) 800 .
Step-by-step Solution
Detailed explanation
The feasible region lies on the origin side of \(y=3\) and \(x+y=8\), and on non-origin side of \(2 x=4\), in the first quadrant.

The corner points of the feasible region are \(\mathrm{A}(2,0), \mathrm{B}(8,0), \mathrm{C}(5,3)\) and \(\mathrm{D}(2,3)\).
\(\mathrm{Z}=100 x+70 y\)
At A(2, 0), Z= 200
At B( 8,0\(), Z=800\)
At C(5,3), Z= 710
At D(2, 3), Z= 410
\(\therefore \quad\) Maximum value of \(Z\) is 800 .

The corner points of the feasible region are \(\mathrm{A}(2,0), \mathrm{B}(8,0), \mathrm{C}(5,3)\) and \(\mathrm{D}(2,3)\).
\(\mathrm{Z}=100 x+70 y\)
At A(2, 0), Z= 200
At B( 8,0\(), Z=800\)
At C(5,3), Z= 710
At D(2, 3), Z= 410
\(\therefore \quad\) Maximum value of \(Z\) is 800 .
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