To solve this problem, let's define the two numbers as $x$ and $y$. We have the following conditions:

1. The product of the two numbers is 108: $x \cdot y = 108$

2. The sum of the first number plus three times the second number should be minimized: $S = x + 3y$

Step-by-Step Solution:

1. Express $x$ in terms of $y$ using the product condition: $x = \frac{108}{y}$

2. Substitute into the sum expression: $S = \frac{108}{y} + 3y$

3. To find the minimum value of $S$, we can use calculus. Take the derivative of $S$ with respect to $y$ and set it to zero: $\frac{dS}{dy} = -\frac{108}{y^2} + 3 = 0$

4. Solve for $y$: $\frac{108}{y^2} = 3$ $y^2 = \frac{108}{3} = 36$ $y = 6$

   Since we're dealing with positive numbers (as the product is positive), we take $y = 6$.

5. Find $x$: $x = \frac{108}{6} = 18$

6. Verify the answer: The pair $(6, 18)$ satisfies the product condition $6 \cdot 18 = 108$. The expression $S = 6 + 3 \cdot 18 = 60$ is the minimum for this combination.

Final Answer:

The correct answer is (c) 6 and 18.

Would you like further details or have any questions?

Here are some related questions you might find interesting:

1. How do you find critical points of a function using derivatives?
2. What is the difference between a minimum and a maximum in calculus?
3. How do you verify if a critical point is a minimum or maximum?
4. What other methods can be used to solve optimization problems without calculus?
5. How would this problem change if the product condition was different?

Tip: When minimizing or maximizing a function, always check if you need to use derivatives, or if simpler methods like completing the square could also work.