$18 \cdot 45$ A rhombus with side length 5, one of its diagonals is no more than 6, and the other is no less than 6, then the maximum value of the sum of the lengths of the two diagonals of this rhombus is
(A) $10 \sqrt{2}$.
(B) 14 .
(C) $5 \sqrt{6}$.
(D) 12 .
(China High School Mathematics League, 1987)
[Solution 1] Let the lengths of the two diagonals of the rhombus be $x$ and $y$. Since the sum of the squares of the diagonals of a parallelogram equals the sum of the squares of its four sides, the problem reduces to: under the conditions
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=100, \\
x \geqslant 6, \\
y \leqslant 6 .
\end{array}\right.
$$

find the maximum value of the function $x+y$.
Consider the family of parallel lines $x+y=c$. When the line $x+y=c$ is tangent to the circle $x^{2}+y^{2}=100$ and the point of tangency is in the first quadrant (as shown in the figure), it is easy to see that $c=10 \sqrt{2}$. This is the maximum value of $x+y$ when only condition (1) is considered.

Since $x$ and $y$ are also subject to the constraints (2) and (3), we move the line $x+y=c$ downward. Clearly, when this line passes through the point $(8,6)$, the maximum value of $x+y$ is 14.
Therefore, the answer is (B).

[Solution 2] Let $x=6+\alpha, y=6-\beta$,
where $\alpha \geqslant 0, \beta \geqslant 0$, substitute the values of $x$ and $y$ into (1),

we get
$$
\begin{array}{c}
(6+\alpha)^{2}+(6-\beta)^{2}=100, \\
(\alpha-\beta)^{2}+12(\alpha-\beta)+2 \alpha \beta-28=0 .
\end{array}
$$

Solving this, we get
$$
\alpha-\beta=-6 \pm \sqrt{64-2 \alpha \beta} \text {. }
$$

When $\beta=0$, and taking the positive sign, we get the maximum value of $\alpha-\beta$ as 2.
$\therefore x+y=12+(\alpha-\beta)$ has a maximum value of 14.
Therefore, the answer is (B).