MHT CET · Maths · Linear Programming
The point, at which the maximum value of \(10 x+6 y\) subject to the constraints \(x+y \leq 12\), \(2 x+y \leq 20, x \geq 0, y \geq 0\) occurs, is
- A \((10,0)\)
- B \((8,4)\)
- C \((0,12)\)
- D \((12,0)\)
Answer & Solution
Correct Answer
(B) \((8,4)\)
Step-by-step Solution
Detailed explanation
The feasible region lies on origin side of \(x+y=12\) and \(2 x+y=20\) and it is in the first quadrant.
The corner point of the feasible region are \(\mathrm{O}(0,0), \mathrm{B}(10,0), \mathrm{C}(8,4), \mathrm{D}(0,12)\).
At \(\mathrm{O}(0,0) \mathrm{z}=10(0)+6(0)=0\)
At B( 10,0\() z=10(10)+6(0)=100\)
\(\operatorname{AtC}(8,4) z=10(8)+6(4)=104\)
At \(\mathrm{D}(0,12) \mathrm{z}=10(0)+6(12)=72\)
\(\therefore \quad\) Maximum value of \(z\) is 104 at it occurs at \(C(8,4)\).
The corner point of the feasible region are \(\mathrm{O}(0,0), \mathrm{B}(10,0), \mathrm{C}(8,4), \mathrm{D}(0,12)\).
At \(\mathrm{O}(0,0) \mathrm{z}=10(0)+6(0)=0\)
At B( 10,0\() z=10(10)+6(0)=100\)
\(\operatorname{AtC}(8,4) z=10(8)+6(4)=104\)
At \(\mathrm{D}(0,12) \mathrm{z}=10(0)+6(12)=72\)
\(\therefore \quad\) Maximum value of \(z\) is 104 at it occurs at \(C(8,4)\).
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