If $ x \plus{} y \equal{} 1$, then the largest value of $ xy$ is:

$ \textbf{(A)}\ 1\qquad 
\textbf{(B)}\ 0.5\qquad 
\textbf{(C)}\ \text{an irrational number about }{0.4}\qquad 
\textbf{(D)}\ 0.25\qquad 
\textbf{(E)}\ 0$
1. Given the equation \( x + y = 1 \), we need to find the maximum value of \( xy \).
2. We can use the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality) which states that for any non-negative real numbers \(a\) and \(b\), the following holds:
   \[
   \frac{a + b}{2} \geq \sqrt{ab}
   \]
3. Applying the AM-GM Inequality to \(x\) and \(y\):
   \[
   \frac{x + y}{2} \geq \sqrt{xy}
   \]
4. Since \( x + y = 1 \), we substitute this into the inequality:
   \[
   \frac{1}{2} \geq \sqrt{xy}
   \]
5. Squaring both sides to eliminate the square root:
   \[
   \left(\frac{1}{2}\right)^2 \geq xy
   \]
   \[
   \frac{1}{4} \geq xy
   \]
6. Therefore, the maximum value of \( xy \) is \( \frac{1}{4} \).
7. To check if this maximum value can be achieved, we set \( x = y \). From \( x + y = 1 \), we get:
   \[
   2x = 1 \implies x = \frac{1}{2}
   \]
   \[
   y = \frac{1}{2}
   \]
8. Substituting \( x = \frac{1}{2} \) and \( y = \frac{1}{2} \) into \( xy \):
   \[
   xy = \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) = \frac{1}{4}
   \]
9. Thus, the maximum value of \( xy \) is indeed \( \frac{1}{4} \).

The final answer is \( \boxed{0.25} \).