Question

If \((2m+n) + (2n+m)=27\), find the maximum value of \((2m-3)\), assuming m and n are positive integers. 
 

Hint:

If an algebra problem seems to have no solution or an unbounded solution (no maximum/minimum), double-check for potential typos. A product might be a sum, or there might be an unstated but implied constraint, such as the variables being positive integers.

Answer 1

Correct Answer: 13

Approach: Collapse the messy left side to a simple linear relation, then push \(m\) as high as the positive-integer constraint allows.

Step 1: Simplify the equation.
\[ (2m + n) + (2n + m) = 27 \implies 3m + 3n = 27 \implies m + n = 9. \]

Step 2: Decide what to maximise.
We want the largest \(2m - 3\), which means the largest \(m\). From \(m = 9 - n\), maximising \(m\) means minimising \(n\).

Step 3: Use the integer constraint.
\(m, n\) are positive integers, so the smallest \(n\) can be is \(1\). Then
\[ m_{\max} = 9 - 1 = 8. \]

Step 4: Evaluate the target.
\[ 2m_{\max} - 3 = 2(8) - 3 = 16 - 3 = 13. \]

Final Answer: \(\boxed{13}\).

Answer 2

Approach: Simplify the given equation first, then treat \( (2m-3) \) as a value to maximise subject to the positive-integer constraint.

Step 1: Expand and combine:
\[ (2m+n)+(2n+m) = 3m+3n = 27 \implies m+n=9. \]
Step 2: Since \( m,n \) are positive integers, \( n\ge1 \), so from \( m=9-n \) we get \( m\le8 \). Also \( m\ge1 \).
Step 3: \( (2m-3) \) increases as \( m \) increases, so it is maximised at the largest allowed \( m \), which is \( m=8 \) (giving \( n=1 \), still a positive integer).
Step 4: Substitute:
\[ 2(8)-3 = \boxed{13}. \]

Final Answer: The maximum value of \( (2m-3) \) is 13.