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Given a rectangular sheet of paper ABCD such that the lengths of AB and AD are respectively 7 and 3 cms.Suppose B' and D' are two points on AB and AD respectively such that if the paper is folded along B'D' then A falls on A' on the side DC. Determine the maximum possible area of the triangle AB'D'.

I discovered quite a few basic facts that everybody can, but cannot actually make any progress. Please help.

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  • $\begingroup$ I sugest " A falls on A' on the side DC. " should be " A falls on A' on the side BC. " $\endgroup$ Jan 7, 2015 at 13:22

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My suggestion is to compute AD' and AB' as a function of AD and $\alpha$. $\Delta AOD' \sim \Delta ADA' \to AD' \times AD = AO\times AA'$

and note that $AA' = \frac{AD}{\cos \alpha}$ and $AO = \frac{1}{2}AA'$

$\therefore AD' = \frac{AA'^2}{2AD}=\frac{AD}{2\cos^2 \alpha}$

$AB' = \frac{AD'}{\tan \alpha} = \frac{AD}{2\cos^2 \alpha} \times \frac{\cos \alpha}{\sin \alpha}=\frac{AD}{\sin 2\alpha}$

$S_{AB'D'} = \frac{1}{2}AB' \times AD' = \frac{AD^2}{2\cos \alpha \sin 2\alpha}$

Since AD is constant, maximizing $S_{AB'D'}$ is to maximize the expression of $\alpha$. You can find the bound of $\alpha$ by moving (1) D' to D and (2) B' to B combining with the assumption 3:7 given. Then, the min or max could be easily obtained.

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Rectangle $ABCD$ is drawn above with vertex $A$ at the origin. Let $DA' = x $ then slope of $AA' = \dfrac{3}{x} $ and therefore, slope of $B'D' = - \dfrac{x}{3} $ and we have that $B'D'$ passes through $( \dfrac{x}{2}, \dfrac{3}{2} )$, therefore, its equation is

$y' = \dfrac{3}{2} - \dfrac{x}{3} (x' - \dfrac{x}{2} ) $

$x'$ intercept is $\dfrac{x}{2} + \dfrac{9}{2x}$ and $y'$ intercept is $\dfrac{3}{2} + \dfrac{x^2}{6} $

We must have the $x'$ intercept in $[0, 7]$, i.e.

$ 0 \le x + \dfrac{9}{x} \le 14 $

So that

$ 0 \le x^2 + 9 \le 14 x $

And finaly $ x^2 - 14 x + 9 \le 0 $, which implies that $x \gt 0.67544 $

And we also have the condition of the $y'$ intercept, namely,

$ AD' = \dfrac{3}{2} + \dfrac{x^2}{6} \le 3 $

From which, we must have $ x \le 3 $

Area of $\triangle AB'D' = [AB'D'] = \dfrac{1}{2} \left(\dfrac{x}{2} + \dfrac{9}{2x} \right) \left(\dfrac{3}{2} + \dfrac{x^2}{6} \right) $

And this simplifies to

$ [AB'D'] = S(x) = \dfrac{1}{2} \left( \dfrac{3}{2} x + \dfrac{1}{12} x^3 + \dfrac{27}{4x} \right) = \dfrac{1}{8} \left( 6 x + \dfrac{x^3}{3} + \dfrac{27}{x} \right)$

Differentiating this expression with respect to $x$ and equating to zero, gives us,

$ 6 + x^2 - 27/x^2 = 0 $

Multiplying through by $x^2$,

$ x^4 + 6 x^2 - 27 = 0 $

$\Rightarrow (x^2 + 9)(x^2 - 3) = 0 $

And therefore, $x = \sqrt{3} $ gives the critical value for the area. To identify it as a local maximum or local minimum, we need to check the sign of second derivative.

$S''(x) = \dfrac{1}{8} ( 2 x + \dfrac{54}{x^3} ) \gt 0 $

So, at $ x = \sqrt{3} $ we have a local minimum.

In addition, we have to check the value of $S(x)$ at the end points $x = 0.67544 $ and $x = 3 $. We have

$S(0.67544) = 5.5161$

$S (\sqrt{3}) = 3.4641$

$S(3) = 4.5$

Therefore, $x = \sqrt{3} $ corresponds to the minimum area of $\triangle AB'D' $ where it is defined.

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