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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?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?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(\left(\frac{16}{5},\frac{27}{5}\right)\)\(\left(\frac{16}{5},\frac{27}{5}\right)\)
Solving the two boundary equations gives \(x=\frac{16}{5}\) and \(y=\frac{27}{5}\). While finding corners, keep only feasible intersections.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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