Finding the maximum area of a right triangle inside a right triangle

Question

So I've encountered a question where it requires me to find the maximum area of a small right angled triangle inside a bigger one. The question stated the dimensions of the big triangle and approved the parallelism of 2 lines in the form below:

Everything in black is given by the question, and otherwise (red) is assumed by me.

I've assumed the line $\overleftrightarrow{eb}$ equal to $x$ and $\overleftrightarrow{db}$ equal to y. Then I proved that $\Delta$ $deb$ is similar to $\Delta$ $abc$ by sharing the same right angle and having the angle $<deb$ corresponding to angle $<acb$, thus:

$$\frac{y}{6} = \frac{x}{8}$$, then $$ y = \frac{3x}{4} $$.

$$\overleftrightarrow{de}$$ would be equal to $$\frac{5x}{4} $$

Now I had to find the height $$\overleftrightarrow{fe}$$ in terms of $$x$$

Since $$\overleftrightarrow{de}$$ is parallel to $$\overleftrightarrow{ac}$$ , angles $$<edf$$ and $$<afd$$ are alternate angles, thus they are equal to each other, and since $$<afd$$ and $$<acb$$ are corresponding angles, $$<edf$$ is equal to $$<acb$$, and both $$\Delta abc, \Delta fed$$ have right angles, then we can infer that $$<efd$$ is equal to $$<cab$$, so both triangles are similar.

Thus: $$\frac{n}{6} = \frac{\frac{5x}{4}}{8}$$ and $$n = \frac{15x}{16} $$.

Now we can find the area of the smaller triangle as a function of $x$.

$$f(x) = 0.5 × \frac{15x}{16} × \frac{5x}{4}$$

But, if I were to take the derivative of that function to find a maximum value, I would end up with a minimum value at $x = 0$, which is utterly irrational. What mistake have I done here?

Answer 1

Hint:let $$ED=m$$ then we get: $$A=\frac{1}{2}mn$$ where $$n=\frac{3}{5}(8-x)$$ and $$m=\sqrt{x^2+\left(\frac{3}{4}x\right)^2}$$ We get then $$A=\frac{3}{8}(8x-x^2)$$ And the maximum we get for $$x=4$$

Answer 2

$ \overline{de} = \frac{5}{4}x $ as you said,

since triangle $ cef$ and triangle $cab$ are similar triangles,

$ \overline{ce} = 8-x $

$ 8-x:n=5:3 $

$5n = 24-3x$

$n = \frac{24-3x}{5}$

$A = \frac{24-3x}{5}\times\frac{5}{4}x\times\frac{1}{2} = \frac{24x-3x^2}{4}\times\frac{1}{2} =3x-\frac{3}{8}x^2$

$\frac{\text{d}A}{\text{d}x} = 3 - \frac{3}{4}x = 0$

$x = 4$

Answer 3

Let $|BC|=a=8$cm, $|AB|=c=6$cm, then $|AC|=b=10$cm.

\begin{align} S_{\triangle FED}(x)&=\tfrac12|DE|\cdot|FE| \\ &=\tfrac12\cdot\frac{x}{\cos\gamma}\cdot(a-x)\cos\alpha \\ &= \tfrac12\cdot\frac{x}{\tfrac{a}{b}}\cdot(a-x)\cdot\tfrac{c}{b} \\ S_{\triangle FED}(x)&=\frac{c}{2a}\,(ax-x^2) . \end{align}

\begin{align} S'_{\triangle FED}(x)&= \frac{c}{2a}\,(a-2x) ,\\ S''_{\triangle FED}(x)&= -\frac{c}{a} <0\quad \forall x , \end{align}

hence $x=\frac{a}2$ provides the maximum

\begin{align} S_{\triangle FED}(x)_{\max}&= S_{\triangle FED}(\tfrac{a}2) =\frac{ac}8 =6\,\mathrm{cm}^2 . \end{align}