Given 3x + y = 1, find the largest value of m for which y/x + 1/y ≤ m always holds.

Step-by-step solution

1. 1

   Reduce to one variable using the constraint. From $3x + y = 1$ we get $y = 1 - 3x$, so the expression becomes a function of $x$ alone:

   $f(x) = \frac{1-3x}{x} + \frac{1}{1-3x} = \frac{1}{x} - 3 + \frac{1}{1-3x}$

   The rewrite $\frac{1-3x}{x} = \frac1x - 3$ is what exposes the two separate poles.

2. 2

   Record the domain. Both denominators must be non-zero:

   $x \ne 0 \quad \text{and} \quad y = 1 - 3x \ne 0 \ \Longrightarrow \ x \ne \tfrac13$

   So the domain is $\mathbb{R} \setminus \left\{0, \tfrac13\right\}$ — and the excluded point $x = \tfrac13$ is exactly where the trouble will be.

3. 3

   Examine the behaviour near the pole x = 1/3. As $x \to \tfrac13^-$ we have $1 - 3x \to 0^+$, hence

   $\frac{1}{1-3x} \to +\infty, \qquad \text{while} \quad \frac1x - 3 \to 3 - 3 = 0$

   The first term blows up and the rest stays bounded, so $f(x) \to +\infty$.

4. 4

   Confirm with explicit values. Taking $x$ closer and closer to $\tfrac13$ from below:

   $x = 0.33 \Rightarrow f = 100.03, \qquad x = 0.333 \Rightarrow f = 1000.003, \qquad x = 0.33333 \Rightarrow f \approx 10^{5}$

   Every target value is exceeded, so the expression has no upper bound. (The same happens as $x \to 0^-$, where $\tfrac1x \to +\infty$ is not the case — there $f \to -\infty$ — but one divergent direction is enough.)

5. 5

   Conclude that no largest m exists. For $\frac{y}{x} + \frac1y \le m$ to hold for all admissible $x, y$, we would need $m \ge \sup f = +\infty$. No real number satisfies that, so:

   $\sup_{3x+y=1} \left(\frac{y}{x} + \frac{1}{y}\right) = +\infty, \qquad \text{no maximum } m \text{ exists}$

   If the intended problem carried extra sign restrictions such as $x > 0,\ y > 0$, note that the pole at $x = \tfrac13^-$ still lies inside that region, so the expression is unbounded there too and the conclusion is unchanged.