असमानताओं \(x+2y \le 8\) और \(2x+y \le 8\) के साथ \(x \ge 0\), \(y \ge 0\) होने पर कौन सा शीर्ष सामान्य सीमा रेखाओं से बनता है?
With \(x+2y \le 8\), \(2x+y \le 8\), \(x \ge 0\), and \(y \ge 0\), which vertex is formed by the two slant boundary lines?
Explanation opens after your attempt
Correct Answer
A. (\left\(\frac{8}{3},\frac{8}{3}\right\))
Concept
Subtracting the two lines gives (x=y), then (3x=8), so \(x=y=\frac{8}{3}\). Find the intersection of slant boundaries separately.
Why this answer is correct
The correct answer is A. (\left\(\frac{8}{3},\frac{8}{3}\right\)). Subtracting the two lines gives (x=y), then (3x=8), so \(x=y=\frac{8}{3}\). Find the intersection of slant boundaries separately.
Exam Tip
दोनों रेखाओं को घटाने पर (x=y), फिर (3x=8) से \(x=y=\frac{8}{3}\)। तिरछी सीमाओं का प्रतिच्छेद अलग से निकालें।
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