असमानताओं \(x+3y\leq 12\), \(2x+y\leq 10\), \(x\geq 0\), \(y\geq 0\) के हल-क्षेत्र में दोनों तिरछी सीमाओं का प्रतिच्छेद कौन सा है?
For the solution region of \(x+3y\leq 12\), \(2x+y\leq 10\), \(x\geq 0\), \(y\geq 0\), what is the intersection of the two slant boundaries?
Explanation opens after your attempt
Correct Answer
A. (\left\(\frac{18}{5},\frac{14}{5}\right\))
Concept
Solving the two equations gives \(x=\frac{18}{5}\) and \(y=\frac{14}{5}\). This is the inner corner on the graph.
Why this answer is correct
The correct answer is A. (\left\(\frac{18}{5},\frac{14}{5}\right\)). Solving the two equations gives \(x=\frac{18}{5}\) and \(y=\frac{14}{5}\). This is the inner corner on the graph.
Exam Tip
दोनों समीकरण हल करने पर \(x=\frac{18}{5}\) और \(y=\frac{14}{5}\) मिलता है। ग्राफ में यही आंतरिक कोना है।
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