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?
Best Answer
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 $$x=4$$