Two numbers add to 6. Let V denote the sum of the first number and the square of twice the second number. What is the smallest possible value for V?

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.

Discover how to minimize V when two numbers add to 6 using a quadratic expression. This problem involves algebraic manipulation and optimization techniques. Suitable for advanced high school students.

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