Find the minimum quotient of two positive numbers x and
y such that 4x + 2y = 100.

We are tasked with finding the minimum quotient \(\frac{x}{y}\) for two positive numbers \(x\) and \(y\) that satisfy the equation:

\[
4x + 2y = 100.
\]

### Step 1: Solve for \(y\) in terms of \(x\)
We can rearrange the given equation to express \(y\) in terms of \(x\). Start by dividing the entire equation by 2 to simplify:

\[
2x + y = 50.
\]

Now, solve for \(y\):

\[
y = 50 - 2x.
\]

### Step 2: Form the quotient
We want to minimize the quotient \(\frac{x}{y}\), where \(y = 50 - 2x\). So, the quotient becomes:

\[
Q(x) = \frac{x}{50 - 2x}.
\]

### Step 3: Minimize the quotient
To find the minimum value of the quotient, we differentiate \(Q(x)\) with respect to \(x\) and set the derivative equal to zero.

First, differentiate using the quotient rule:

\[
Q'(x) = \frac{(50 - 2x)(1) - x(-2)}{(50 - 2x)^2} = \frac{50 - 2x + 2x}{(50 - 2x)^2} = \frac{50}{(50 - 2x)^2}.
\]

Set \(Q'(x) = 0\):

\[
\frac{50}{(50 - 2x)^2} = 0.
\]

Since the numerator is 50 (which is non-zero), this equation has no solution. This implies that the quotient has no critical points from the first derivative.

### Step 4: Boundary considerations
Since the derivative does not yield a critical point, we need to consider boundary values where \(x\) and \(y\) are positive.

From the constraint \(4x + 2y = 100\), or equivalently \(2x + y = 50\), we know that \(0 < x < 25\) (because if \(x = 25\), \(y = 0\), which is not allowed since \(y\) must be positive).

Thus, the range of \(x\) is \(0 < x < 25\), and the range of \(y = 50 - 2x\) is \(0 < y < 50\).

### Step 5: Evaluate at the boundary
Evaluate the quotient \(Q(x)\) at the boundaries of \(x\):

- When \(x \to 0^+\), we have \(y \to 50\), so \(\frac{x}{y} = \frac{0}{50} = 0\).
- When \(x \to 25^-\), we have \(y \to 0^+\), so \(\frac{x}{y} \to \infty\).

Thus, the minimum quotient \(\frac{x}{y}\) occurs as \(x \to 0^+\), and the minimum value is 0.

### Conclusion:
The minimum quotient \(\frac{x}{y}\) is 0, which occurs as \(x \to 0^+\).

---

Do you want more details or have any questions? Here are some related questions:

1. How do you apply the quotient rule to different types of functions?
2. Can the quotient of two numbers be negative in this scenario?
3. What happens if we change the equation to \(4x + 2y = k\) for a different constant \(k\)?
4. How do boundary conditions affect the solution of optimization problems?
5. Can we use another method (like Lagrange multipliers) to solve this problem?

**Tip:** In optimization problems, always check the boundaries if the derivative test fails to find critical points.

Find the minimum quotient of two positive numbers x and y such that 4x + 2y = 100.

Discover how to find the minimum quotient x/y for two positive numbers x and y that satisfy 4x + 2y = 100. This optimization problem involves using the quotient rule and boundary condition analysis. Detailed steps are provided for students in Grades 10-12.

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