असमानताओं \(x+2y\leq 14\), \(3x+y\leq 15\), \(x\geq 0\), \(y\geq 0\) के हल-क्षेत्र का वह कोना कौन सा है जहाँ दोनों तिरछी सीमाएं मिलती हैं?
For the solution region of \(x+2y\leq 14\), \(3x+y\leq 15\), \(x\geq 0\), and \(y\geq 0\), which corner is formed by the two slant boundaries?
Explanation opens after your attempt
Correct Answer
A. (\left\(\frac{16}{5},\frac{27}{5}\right\))
Concept
Solving the two boundary equations gives \(x=\frac{16}{5}\) and \(y=\frac{27}{5}\). While finding corners, keep only feasible intersections.
Why this answer is correct
The correct answer is A. (\left\(\frac{16}{5},\frac{27}{5}\right\)). Solving the two boundary equations gives \(x=\frac{16}{5}\) and \(y=\frac{27}{5}\). While finding corners, keep only feasible intersections.
Exam Tip
दोनों सीमा समीकरण हल करने पर \(x=\frac{16}{5}\) और \(y=\frac{27}{5}\) मिलता है। कोना निकालते समय केवल वैध प्रतिच्छेद लें।
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