Let x1, x2, x3 ≥ 0 and x1 + x2 + x3 = 1. If the maximum value of

(x1 + 3x2 + 5x3)(x1 +x2/3 +x3/5)=m/n

(where m and n are co-primes).

Then find the value of m + n.

Given the problem, we are tasked with finding the maximum value of the expression:

\[
f(x_1, x_2, x_3) = \left(x_1 + 3x_2 + 5x_3\right)\left(x_1 + \frac{x_2}{3} + \frac{x_3}{5}\right)
\]

subject to the constraints \(x_1 + x_2 + x_3 = 1\) and \(x_1, x_2, x_3 \geq 0\).

Let's first rewrite the function to simplify it:
\[
f(x_1, x_2, x_3) = \left(x_1 + 3x_2 + 5x_3\right)\left(x_1 + \frac{x_2}{3} + \frac{x_3}{5}\right)
\]

We will evaluate \(f(x_1, x_2, x_3)\) at some specific points where \(x_1, x_2, x_3\) are non-negative and add up to 1.

### 1. Case: \(x_1 = 1\), \(x_2 = 0\), \(x_3 = 0\)
\[
f(1, 0, 0) = \left(1 + 3 \times 0 + 5 \times 0\right) \left(1 + \frac{0}{3} + \frac{0}{5}\right) = 1 \times 1 = 1
\]

### 2. Case: \(x_1 = 0\), \(x_2 = 1\), \(x_3 = 0\)
\[
f(0, 1, 0) = \left(0 + 3 \times 1 + 5 \times 0\right) \left(0 + \frac{1}{3} + \frac{0}{5}\right) = 3 \times \frac{1}{3} = 1
\]

### 3. Case: \(x_1 = 0\), \(x_2 = 0\), \(x_3 = 1\)
\[
f(0, 0, 1) = \left(0 + 3 \times 0 + 5 \times 1\right) \left(0 + \frac{0}{3} + \frac{1}{5}\right) = 5 \times \frac{1}{5} = 1
\]

### 4. Case: \(x_1 = 0.5\), \(x_2 = 0.5\), \(x_3 = 0\)
\[
f\left(0.5, 0.5, 0\right) = \left(0.5 + 3 \times 0.5 + 5 \times 0\right) \left(0.5 + \frac{0.5}{3} + \frac{0}{5}\right)
\]
\[
= \left(0.5 + 1.5\right) \left(0.5 + \frac{1}{6}\right) = 2 \times \frac{4}{6} = 2 \times \frac{2}{3} = \frac{4}{3} \approx 1.333
\]

### 5. Case: \(x_1 = \frac{1}{3}\), \(x_2 = \frac{1}{3}\), \(x_3 = \frac{1}{3}\)
\[
f\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right) = \left(\frac{1}{3} + 3 \times \frac{1}{3} + 5 \times \frac{1}{3}\right) \left(\frac{1}{3} + \frac{\frac{1}{3}}{3} + \frac{\frac{1}{3}}{5}\right)
\]
\[
= \left(\frac{1}{3} + 1 + \frac{5}{3}\right) \left(\frac{1}{3} + \frac{1}{9} + \frac{1}{15}\right) = \left(\frac{9}{3}\right) \left(\frac{15 + 5 + 3}{45}\right)
\]
\[
= 3 \times \frac{23}{45} = \frac{69}{45} = \frac{23}{15} \approx 1.533
\]

The maximum value occurs at \(x_1 = 0.5\), \(x_2 = 0.5\), \(x_3 = 0\), which gives:
\[
f(0.5, 0.5, 0) = \frac{4}{3}
\]

Thus, \( m = 4 \) and \( n = 3 \). Therefore, the value of \( m + n \) is:
\[
\boxed{7}
\]

Would you like further details or have any questions?

### Follow-up Questions:
1. How does the function behave as \(x_1\), \(x_2\), \(x_3\) vary?
2. Can we derive a general strategy for maximizing functions like this?
3. What role do the constraints play in maximizing the function?
4. How do you handle cases where \(m\) and \(n\) are not co-prime?
5. What if the function involved different coefficients?

**Tip:** When solving optimization problems, always evaluate the function at critical points and boundary values based on given constraints.

Explore how to maximize the quadratic expression (x1 + 3x2 + 5x3)(x1 + x2/3 + x3/5) under given constraints. Detailed step-by-step solution with evaluation at critical points. Suitable for advanced high school students and beyond.

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