Let's denote the two numbers as $x$ and $y$. We are given:

$x + y = 6$

We want to minimize $V$, which is defined as:

$V = x + (2y)^2$

First, simplify the expression for $V$:

$V = x + 4y^2$

Since $x + y = 6$, we can express $x$ in terms of $y$:

$x = 6 - y$

Substitute this into the expression for $V$:

$V = (6 - y) + 4y^2$

Simplifying, we get:

$V = 6 - y + 4y^2$

To find the minimum value of $V$, we'll take the derivative of $V$ with respect to $y$ and set it equal to zero:

$\frac{dV}{dy} = -1 + 8y$

Setting the derivative equal to zero:

$-1 + 8y = 0$

Solving for $y$:

$8y = 1 \quad \Rightarrow \quad y = \frac{1}{8}$

Next, substitute $y = \frac{1}{8}$ back into the expression for $V$ to find the minimum value:

$V = 6 - \frac{1}{8} + 4\left(\frac{1}{8}\right)^2$

Simplify this expression:

$V = 6 - \frac{1}{8} + 4 \times \frac{1}{64}$

$V = 6 - \frac{1}{8} + \frac{1}{16}$

$V = 6 - \frac{2}{16} + \frac{1}{16} = 6 - \frac{1}{16}$

$V = \frac{96}{16} - \frac{1}{16} = \frac{95}{16} = 5.9375$

So, the smallest possible value for $V$ is $\boxed{5.9375}$.

Would you like any more details or have any questions?

Here are some related questions:

1. How does changing the coefficient of $y^2$ affect the minimum value of $V$?
2. What does the second derivative tell us about the nature of the minimum?
3. How would the problem change if the numbers added to a different sum instead of 6?
4. Can you find a maximum value for $V$ given these constraints?
5. How can this method be extended to minimize other similar expressions?

Tip: When dealing with quadratic functions, the minimum or maximum value can often be found efficiently by taking the derivative and solving for critical points.