KCET · Maths · Linear Programming
The feasible region of an LPP is shown in the figure. If \( z=3 x+9 y \), then the minimum value of
z occurs at
- A \( (5,5) \)
- B \( (0,10) \)
- C \( (0,20) \)
- D \((15,15) \)
Answer & Solution
Correct Answer
(A) \( (5,5) \)
Step-by-step Solution
Detailed explanation
Given that, \( z=3 x+9 y \)
The least point is \( (5,5) \)
\[
\begin{array}{l}
\text { So, } z(5,5)=3(5)+9(5) \\
=15+45=60
\end{array}
\]
Now take point \( (0,10) \).
So, \( z(0,10)=3(0)+9(10)=90 \)
Therefore, minimum value of \( z \) is at \( (5,5) \)
The least point is \( (5,5) \)
\[
\begin{array}{l}
\text { So, } z(5,5)=3(5)+9(5) \\
=15+45=60
\end{array}
\]
Now take point \( (0,10) \).
So, \( z(0,10)=3(0)+9(10)=90 \)
Therefore, minimum value of \( z \) is at \( (5,5) \)
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