What is the maximum value of ac in the given equations?

[anemone]

Consider the following equations:

$a^2+b^2=16$

$c^2+d^2=25$

$ad-bc=20$

where $a,\,b,\,c,\,d \in R$

Find the maximum value of $ac$.

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[anemone]

Congratulations to the following members for their correct solutions:

1. laura123
2. lfdahl
3. kaliprasad

Solution from laura123:

Spoiler

Let us consider the points $A(a,b)$ and $B(c,d)$.
Since $a^2+b^2=16$ and $c^2+d^2=25$ it follows that $A$ belongs to a circle with centre $(0,0)$ and radius 4 and $B$ belongs to a circle with centre $O(0,0)$ and radius 5 , as shown in the following figure:

$ad-bc=\overline{OA}\cdot\overline{OB}\cdot\sin\theta=4\cdot 5\cdot\sin\theta=20\sin\theta$.
Since $ad-bc=20$ we have $20\sin\theta=20$ i.e. $\theta=\dfrac{\pi}{2}$.
Let $\alpha$ be the angle that $OA$ makes with the x-axis ($0\leq \alpha<2\pi$) as shown in the following figure:

$ac=\overline{OA}\cos\alpha\cdot\overline{OB}\cos\left(\alpha+\dfrac{\pi}{2}\right)=4\cos\alpha\cdot 5\cos\left(\alpha+\dfrac{\pi}{2}\right)=$
$=20\cos\alpha\cos\left(\alpha+\dfrac{\pi}{2}\right)=-20\cos\alpha\sin\alpha=-10\sin 2\alpha$

Therefore, the maximum value of $ac$ is $10$.

Solution from lfdahl:

Spoiler

Define the two dimensional real valued vectors: $\boldsymbol{v_1}= \begin{pmatrix} a\\b\end{pmatrix}$ and $\boldsymbol{v_2}= \begin{pmatrix}c\\d\end{pmatrix}$.

Then the given equations can be expressed as:

$\left \| \boldsymbol{v_1} \right \|^2 =v_1^2= 16$ and $\left \| \boldsymbol{v_2} \right \|^2 =v_2^2= 25$ and $det(\boldsymbol{v_1},\boldsymbol{v_2})=20$.

The last equation expresses the area of the parallelogram spanned by $\boldsymbol{v_1} $ and $\boldsymbol{v_2} $.

With $v_1=4$ and $v_2=5$ this is only possible, if \[\boldsymbol{v_1} \perp \boldsymbol{v_2}\]
- for all possible choices of orthogonal vectors $\boldsymbol{v_1} $ and $\boldsymbol{v_2} $ in the Cartesian plane.

If we let
\[\boldsymbol{v_1}=v_1\begin{pmatrix} cos\alpha \\ sin\alpha \end{pmatrix}, \: \: \: \boldsymbol{v_2}=v_2\begin{pmatrix} -sin\alpha \\ cos\alpha \end{pmatrix},\: \: \: \: 0\leq \alpha < 2\pi\]
then
\[max\left \{ ac \right \}=max\left \{ -v_1v_2cos\alpha sin\alpha \right \}=-\frac{v_1v_2}{2}min\left \{ sin2\alpha \right \}=\frac{v_1v_2}{2}=10.\]

Solution from kaliprasad:

Spoiler

Without loss of generality we can choose
$a=4\sin\,t$
$b=4\cos\,t$
$c=5\sin\,p$
$d=5\cos\,p$

So we get $ad-bc= 20\sin\, t \cos\, p - 20\sin\, p \cos\, t = 20\sin (t-p) = 20$
or $\sin(t-p) = 1$
so $t= p+ \dfrac{\pi}{2}$
Hence
$ac = 20 \ sin \, t \ sin \ p$
= $20 \ sin\, p +\dfrac{\pi}{2} \ sin\, p$
= $-20 \cos \, p \sin\, p$
= $-10 \sin 2p$

Clearly the largest value is 10 and smallest -10.