<s>[INST] Any purely geometric solution to this problem? 
What is the largest possible area of a rectangle(in square units) inscribed in the triangle shown in the picture above?
 [/INST] Let $\triangle ABC$ be the given triangle with $AB = 10$, $AC = 17$, and $BC = 21$. Choose an arbitrary point $M$ on the side $AB$, and let $x = AM$. Let $N$ be the point on $AC$ such that $MN \parallel  BC$, and points $P$, and $Q$ on $BC$ such that $NP \perp BC$, and $MQ \perp BC$. Thus $MNPQ$ is a rectangle. Let $h_a$ be the length of the altitude from $A$. Note that $h_a$ is constant. The we have the followings proportions:
$\dfrac{BM}{BA} = \dfrac{MQ}{h_a} \to \dfrac{10-x}{10} = \dfrac{MQ}{h_a} \to MQ = \dfrac{h_a}{10}\cdot (10-x)$. Also:
$\dfrac{AM}{AB} = \dfrac{MN}{BC} \to \dfrac{x}{10} = \dfrac{MN}{21} \to MN = \dfrac{21}{10}\cdot x$. 
Let $S(x)$ be the area of the rectangle $MNPQ$, then:
$S(x)  = MN\cdot MQ = \dfrac{21h_a}{100}\cdot x(10 - x)$. Since:
$x + (10 - x) = 10$, a constant, $S$ achieves a maximum value when $x = 10 - x$, or $x = 5$.
From this we can solve for $MN =\dfrac{21}{2}$, and $MQ = \dfrac{h_a}{2}$. Thus:
$S_{max} = \dfrac{21h_a}{4} = \dfrac{S_{\triangle ABC}}{2}$
Note: $h_a$ can be calculated using Heron formula.
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