The problem asks to find two positive numbers whose product is 192, and where the sum of the first number plus three times the second number is minimized.

Let's call the two numbers $x$ and $y$.

Step 1: Define the relationship

We know: $x \cdot y = 192$ This gives us the equation: $y = \frac{192}{x}$

Step 2: Set up the function to minimize

We are asked to minimize the sum of the first number plus three times the second. This gives us the function: $S(x, y) = x + 3y$ Substituting $y = \frac{192}{x}$ into this equation, we get: $S(x) = x + 3\left(\frac{192}{x}\right) = x + \frac{576}{x}$

Step 3: Minimize the function

To find the value of $x$ that minimizes this function, we take the derivative of $S(x)$ with respect to $x$ and set it equal to zero: $S'(x) = 1 - \frac{576}{x^2}$ Setting the derivative equal to zero to find the critical points: $1 - \frac{576}{x^2} = 0$ $x^2 = 576$ $x = \sqrt{576} = 24$

Step 4: Find the corresponding $y$

Since $x = 24$, we can find $y$ using the equation $x \cdot y = 192$: $24 \cdot y = 192 \quad \Rightarrow \quad y = \frac{192}{24} = 8$

Conclusion:

The two numbers are $\mathbf{24}$ and $\mathbf{8}$.

Would you like a more detailed breakdown of the steps? Here are some related questions:

1. What if the product was a different number, like 150? How would the method change?
2. How do you check if the result really minimizes the sum?
3. Can we use graphical methods to find the minimum instead of calculus?
4. Why do we take the derivative to find minimum values?
5. What happens if we change the second number's coefficient (currently 3) to another number?

Tip: When minimizing or maximizing a function, always check if you are working with constraints, as it can affect your result.