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Inside $\triangle EFG$ with surface of $9 \ cm^2$ there is a rectangle $ABCD$ shuch as $AB=2CD$.
I need to find the maximal length of BC.
I've tried putting $BC=y$ and $x$ to several other stuff in order to find $y(x)$ and then solving $y'(x)=0$ in order to maximize $x$, but I'm getting equations that makes is really hard to solve for $y$.
Any ideas maybe? Thanks

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    Do you mean the maximal length of $BC$ over all rectangles contained in _any_ triangle $\triangle EFG$ of area $9\text{cm}^2$? If so, I suggest re-casting the problem, e.g. "Let $R$ be a rectangle with dimensions $2\text{cm} \times 1\text{cm}$. What is the smallest possible area of a triangle containing $R$?" – 2017-01-24
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    And I guess it is $AB=2\cdot BC$, otherwise it is not a rectangle. – 2017-01-24

2 Answers 2

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First we have $\Delta EAB$ and $\Delta EFG$ are similars. Let's call $GF=a$ and the height of $\Delta EFG$ equal to $h$. Then:

$$\frac{h-x}{h}=\frac{2x}{a}$$

Once the area of $\Delta EFG$ is $9$ then $a\cdot h=18 \to h=18/a$ we get

$$x=\frac{18a}{36+a^2}=f(a)$$

For the maximum $x$ we have $f'(a)=0 \to a=6$ and then

$$x=\frac{3}{2}$$

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    You are very welcome. – 2017-01-24
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The dual problem is to find the smallest triangle circumscribed to a $2\times 1$ rectangle, but if triangle is circumscribed to a $2\times 1$ rectangle, by "paper folding" we get that the area of the triangle is at least twice the area of the rectangle. It follows that the largest inscribed rectangle with such proportions inscribed in a triangle with area $9$ has area $\frac{9}{2}$, hence shortest side with length $\color{red}{\frac{3}{2}}$.

enter image description here

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    Thank you, it was very educational and interesting solution, but still I'm looking for more of an analytical solution. Getting $y'(x)=0$ using geometrical ways. Using triangle similarity or something alike. – 2017-01-24