Problem source

Show that, if the lengths of the diagonals of a parallelogram are specified, then the parallogram has maximum area when the diagonals are perpendicular. Show also that the area of a parallelogram is less than or equal to half the square of the length of its longer diagonal. 
The set $A$ of points $(x,y)$ is given by 
\begin{alignat*}{1}
\left|a_{1}x+b_{1}y-c_{1}\right| & \leqslant\delta,\\
\left|a_{2}x+b_{2}y-c_{2}\right| & \leqslant\delta,
\end{alignat*}
 with $a_{1}b_{2}\neq a_{2}b_{1}.$ Sketch this set and show that it is possible to find $(x_{1},y_{1}),(x_{2},y_{2})\in A$ with 
\[
(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}\geqslant\frac{8\delta^{2}}{\left|a_{1}b_{2}-a_{2}b_{1}\right|}.
\]

Solution source

In a parallelogram the diagonals meet at their mid points. Fixing one diagonal, we can look at the two triangles formed by the other diagonal. Suppose the angle between them is $\theta$. Then the area of the triangles will be $\frac12 \frac{l_1}{2} \frac{l_2}2 \sin \theta+\frac12 \frac{l_1}{2} \frac{l_2}2 \sin (\pi -\theta) = \frac{l_1l_2}{4} \sin \theta$. This will be true on both sides. Therefore we can maximise this area by setting $\theta = \frac{\pi}{2}$.

\begin{center}
    \begin{tikzpicture}[scale=2]
        \coordinate (A) at (-1,0.5);
        \coordinate (B) at (1,-0.5);

        \coordinate (C) at (-1,-0.5);
        \coordinate (D) at (1,0.5);

        \def\d{0.25};

        \draw (A) -- (B);
        \draw (C) -- (D);

        \def\ka{1/sqrt(3)};
        \def\kb{sqrt(2)/sqrt(3)};

        \draw[dashed] ($(A)+{\d}*({\ka},{\kb})$) --  ($(B)+{\d}*({\ka},{\kb})$);
        \draw[dashed] ($(A)-{\d}*({\ka},{\kb})$) --  ($(B)-{\d}*({\ka},{\kb})$);

        \filldraw[color=blue, opacity=0.2] ($(A)+{\d}*({\ka},{\kb})$) --  ($(B)+{\d}*({\ka},{\kb})$) --  ($(B)-{\d}*({\ka},{\kb})$) --  ($(A)-{\d}*({\ka},{\kb})$);

        \def\kc{1/sqrt(3)};
        \def\kd{-sqrt(2)/sqrt(3)};


        \draw[dashed] ($(C)+{\d}*({\kc},{\kd})$) --  ($(D)+{\d}*({\kc},{\kd})$);
        \draw[dashed] ($(C)-{\d}*({\kc},{\kd})$) --  ($(D)-{\d}*({\kc},{\kd})$);

        \filldraw[color=blue, opacity=0.2]  ($(C)+{\d}*({\kc},{\kd})$) --  ($(D)+{\d}*({\kc},{\kd})$) -- ($(D)-{\d}*({\kc},{\kd})$) -- ($(C)-{\d}*({\kc},{\kd})$);
    \end{tikzpicture}
\end{center}

Consider the (darker) shaded area. This is our set $A$.

The area of the set is indifferent to a parallel shift in the lines, so without loss of generality, we can consider $c_1 = 0, c_2 = 0$, so our lines meet at the origin.

Now also consider the linear transformation $\begin{pmatrix} a_1  & b_1 \\ a_2 & b_2 \end{pmatrix}^{-1}$ which takes the coordinate axes to these lines. This will take the square $[-\delta, \delta] \times [-\delta, \delta]$ which has area $4\delta^2$ to our square, which will have area $\frac{4 \delta^2}{ |a_1b_2 - a_2b_1|}$. 

If we consider the length of the two diagonals of this area, $l_1, l_2$ we know that $\frac{l_1l_2}2 \sin \theta = \frac{4 \delta^2}{|a_1b_2 - a_2b_1|}$, if we consider the larger of $l_1$ and $l_2$ (wlog $l_1$) we must have that $\frac{l_1^2}{2} \geq  \frac{4 \delta^2}{|a_1b_2 - a_2b_1|}$ and so points on opposite ends of the diagonal will satisfy the inequality in the question.