<s>[INST] The largest area of the rectangle inscribed into an acute triangle 
A triangle of base $b$ and height $h$ has acute base angles. A rectangle is inscribed in the triangle with one side on the base of the triangle. Show that the largest such rectangle has base $b/2$ and height $h/2$, so that its area is one-half the area of the triangle.

So, we have a triangle as it is shown in the picture (Sorry for the awful quality). We have that $bh/2 = A$, where $A$ is the area of the triangle. We have $y/(b-x-z) = h/(b-x-z+x_1)$ from where $y = \frac{h(b-x-z)}{b-x-z+x_1}$ and $y/z = h/(x_2+z)$, from where $y = \frac{zh}{x_2+z}$. So $\frac{zh}{x_2+z} = \frac{h(b-x-z)}{b-x-z+x_1}$. From this I get $zhx = x_2h(b-x)$. Can this be useful? If so, what should be done next?
 [/INST] The solution for the general case requires a bit more effort than the right triangle case. We will be using a lot of "Basic Probotionality theorem for triangles"

$$\frac{y}{h} = \frac{b_1 - x_1}{b_1}$$
Similarly, 
$$\frac{y}{h} = \frac{b_2 - x_2}{b_2}$$
Giving us,
$$\frac{x_1}{b_1} = \frac{x_2}{b_2}$$
Using Compedendo dividendo,we get
$$\frac{x_1}{b_1} = \frac{x_2}{b_2} = \frac{x_1 + x_2}{b_1 + b_2} = \frac{x}{b}$$
Now, 
$$\frac{y}{h} = 1 - \frac{x_1}{b_1}$$
Hence, 
$$\frac{y}{h} = 1 - \frac{x}{b}$$
Multiplying with $x$ on both sides we get,
$$ xy = hx - \frac{hx^2}{b}$$
$$ A = hx - \frac{hx^2}{b}$$
$$ \frac{dA}{dx} = h - \frac{2hx}{b}$$
For maxima this has to be equal to zero.
$$ h = \frac{2hx}{b}$$
$$ x = \frac{b}{2}$$
Also, 
$$y = \frac{h}{2}$$
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