MHT CET · Maths · Linear Programming
The maximum value of the objective function \(\mathrm{z}=4 x+6 y\) subject to \(3 x+2 y \leq 12, x+y \geq 4, x\), \(y \geq 0\) is
- A 24
- B 46
- C 56
- D 36
Answer & Solution
Correct Answer
(D) 36
Step-by-step Solution
Detailed explanation
The corner points of feasible region are \(\mathrm{A}(4,0)\), \(\mathrm{B}(0,4), \mathrm{C}(0,6)\)

At \(A(4,0), Z=4(4)+6(0)=16\)
At B( 0,4\(), Z=4(0)+6(4)=24\)
At \(\mathrm{C}(0,6), \mathrm{Z}=4(0)+6(6)=36\)
\(\therefore \quad \mathrm{Z}\) has maximum value at \(\mathrm{C}(0,6)\) which is 36 .

At \(A(4,0), Z=4(4)+6(0)=16\)
At B( 0,4\(), Z=4(0)+6(4)=24\)
At \(\mathrm{C}(0,6), \mathrm{Z}=4(0)+6(6)=36\)
\(\therefore \quad \mathrm{Z}\) has maximum value at \(\mathrm{C}(0,6)\) which is 36 .
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