0 reads0 ratings0 helpful
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?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?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(\left(\frac{10}{3},\frac{10}{3}\right)\)\(\left(\frac{10}{3},\frac{10}{3}\right)\)
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.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।
Student feedback
Was this question useful?
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Difficulty
Easy0%
Medium0%
Hard0%
Was the explanation clear?
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
