असमानताओं \(2x+y \le 10\), \(x+2y \le 10\), \(x \ge 0\), \(y \ge 0\) के क्षेत्र में अधिकतम संभावित (x+y) किस शीर्ष पर मिलेगा?
For the region \(2x+y \le 10\), \(x+2y \le 10\), \(x \ge 0\), \(y \ge 0\), at which vertex is the maximum possible (x+y) obtained?
Explanation opens after your attempt
C. (\left\(\frac{10}{3},\frac{10}{3}\right\))
Concept
The slant lines intersect at (\left\(\frac{10}{3},\frac{10}{3}\right\)), where \(x+y=\frac{20}{3}\). Check linear expressions at vertices for maxima.
Why this answer is correct
The correct answer is C. (\left\(\frac{10}{3},\frac{10}{3}\right\)). The slant lines intersect at (\left\(\frac{10}{3},\frac{10}{3}\right\)), where \(x+y=\frac{20}{3}\). Check linear expressions at vertices for maxima.
Exam Tip
दो तिरछी रेखाओं का प्रतिच्छेद (\left\(\frac{10}{3},\frac{10}{3}\right\)) है और वहां \(x+y=\frac{20}{3}\) मिलता है। रैखिक अभिव्यक्ति का अधिकतम कोनों पर जांचें।
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