KCET · Maths · Probability
The feasible region of an LPP is shown in the figure. If \(z=11 x+7 y\), then the maximum value of \(Z\) occurs at
- A \((0,5)\)
- B \((3,3)\)
- C \((5,0)\)
- D \((3,2)\)
Answer & Solution
Correct Answer
(D) \((3,2)\)
Step-by-step Solution
Detailed explanation
Given, maximize \(z=11 x+7 y\)
Intersecting point of lines \(x+y=5\) and \(x+3 y=9\) is \((3,2)\).
\(\therefore\) Corner point is \(B(3,2)\)
For corner points of the feasible region
We put, \(x=0\) in \(x+3 y=9\)
\(y=3\)
\(\Rightarrow\) corner point is \(A(0,3)\)
and put \(x=0\) in \(x+y=5\), we get \(y=5\)
\(\Rightarrow\) corner point is \(C(0,5)\)
The values of \(z\) at these corner points are as follows
Therefore, maximum value of \(z\) is 47 at \((3,2)\).
Intersecting point of lines \(x+y=5\) and \(x+3 y=9\) is \((3,2)\).
\(\therefore\) Corner point is \(B(3,2)\)
For corner points of the feasible region
We put, \(x=0\) in \(x+3 y=9\)
\(y=3\)
\(\Rightarrow\) corner point is \(A(0,3)\)
and put \(x=0\) in \(x+y=5\), we get \(y=5\)
\(\Rightarrow\) corner point is \(C(0,5)\)
The values of \(z\) at these corner points are as follows
Therefore, maximum value of \(z\) is 47 at \((3,2)\).
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