यदि \(x\ge1\), \(y\ge2\), \(x+y\le8\) हैं, तो हल क्षेत्र के शीर्ष कौन से हैं?

If \(x\ge1\), \(y\ge2\), \(x+y\le8\), what are the vertices of the solution region?

Explanation opens after your attempt
Correct Answer

A. ( (1,2) ), ( (6,2) ), ( (1,7) )

Step 1

Concept

The boundaries (x=1), (y=2), and (x+y=8) give ( (1,2) ), ( (6,2) ), and ( (1,7) ). Do not treat shifted restrictions as simple \(x\ge0\), \(y\ge0\).

Step 2

Why this answer is correct

The correct answer is A. ( (1,2) ), ( (6,2) ), ( (1,7) ). The boundaries (x=1), (y=2), and (x+y=8) give ( (1,2) ), ( (6,2) ), and ( (1,7) ). Do not treat shifted restrictions as simple \(x\ge0\), \(y\ge0\).

Step 3

Exam Tip

सीमाएं (x=1), (y=2) और (x+y=8) से ( (1,2) ), ( (6,2) ), ( (1,7) ) मिलते हैं। बदले हुए अक्षीय प्रतिबंधों को साधारण \(x\ge0\), \(y\ge0\) न मानें।

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Mathematics Answer, Explanation and Revision Hints

यदि \(x\ge1\), \(y\ge2\), \(x+y\le8\) हैं, तो हल क्षेत्र के शीर्ष कौन से हैं? / If \(x\ge1\), \(y\ge2\), \(x+y\le8\), what are the vertices of the solution region?

Correct Answer: A. ( (1,2) ), ( (6,2) ), ( (1,7) ). Explanation: सीमाएं (x=1), (y=2) और (x+y=8) से ( (1,2) ), ( (6,2) ), ( (1,7) ) मिलते हैं। बदले हुए अक्षीय प्रतिबंधों को साधारण \(x\ge0\), \(y\ge0\) न मानें। / The boundaries (x=1), (y=2), and (x+y=8) give ( (1,2) ), ( (6,2) ), and ( (1,7) ). Do not treat shifted restrictions as simple \(x\ge0\), \(y\ge0\).

Which concept should I revise for this Mathematics MCQ?

The boundaries (x=1), (y=2), and (x+y=8) give ( (1,2) ), ( (6,2) ), and ( (1,7) ). Do not treat shifted restrictions as simple \(x\ge0\), \(y\ge0\).

What exam hint can help solve this Mathematics question?

सीमाएं (x=1), (y=2) और (x+y=8) से ( (1,2) ), ( (6,2) ), ( (1,7) ) मिलते हैं। बदले हुए अक्षीय प्रतिबंधों को साधारण \(x\ge0\), \(y\ge0\) न मानें।