KCET · Maths · Three Dimensional Geometry
If \( x+y \leq 2, x \geq 0, y \geq 0 \) the point at which maximum value of \( 3 x+2 y \) attained will be
- A \( (0,0) \)
- B \( \left(\frac{1}{2}, \frac{1}{2}\right) \)
- C \( (0,2) \)
- D \( (2,0) \)
Answer & Solution
Correct Answer
(D) \( (2,0) \)
Step-by-step Solution
Detailed explanation
Given that \(x+y \leq 2 \rightarrow(1)\)
\(x \geq 0 . y \geq 0 \rightarrow(2)\)
Corner points are \((0,0),(2,0),(0,2)\)
Maximum of \(3 x+2 y\) is at \((2,0)\)
\[
\begin{array}{l}
\text { Given that } x+y \leq 2 \rightarrow(1) \\
x \geq 0 . y \geq 0 \rightarrow(2) \\
\text { Corner points are }(0,0),(2,0),(0,2) \\
\text { Maximum of } 3 x+2 y \text { is at }(2,0)
\end{array}
\]
\(x \geq 0 . y \geq 0 \rightarrow(2)\)
Corner points are \((0,0),(2,0),(0,2)\)
Maximum of \(3 x+2 y\) is at \((2,0)\)
\[
\begin{array}{l}
\text { Given that } x+y \leq 2 \rightarrow(1) \\
x \geq 0 . y \geq 0 \rightarrow(2) \\
\text { Corner points are }(0,0),(2,0),(0,2) \\
\text { Maximum of } 3 x+2 y \text { is at }(2,0)
\end{array}
\]
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