a rectangle is to be inscribed in a right triangle having sides of lengty 6 in, 8 in, and 10 in. find the dimension of the rectangle with greatest area..
tq~
Hi redcherry,
If you draw the right-triangle, with an arbitrary inscribed rectangle,
that has sides against the base and vertical side, settling in the
right-angled corner.
label the base of the triangle 6, height 8, hypotenuse 10.
The rectangle base is x and the rectangle height y.
Now, observe the smaller identical triangle at the bottom corner.
It has base (6-x) and height y.
Hence we can write an equation
$\displaystyle tan\theta=\frac{y}{6-x}=\frac{8}{6}\ \Rightarrow\ 6y=8(6-x)\ \Rightarrow\ y=\frac{8}{6}(6-x)$
Rectangle area is
$\displaystyle xy=\frac{8x}{6}(6-x)$
in terms of x only, allowing us to differentiate the area and equate the derivative to zero to find the rectangle of maximum area
$\displaystyle \frac{8}{6}\frac{d}{dx}\left[x(6-x)\right]=\frac{8}{6}\frac{d}{dx}\left(6x-x^2\right)$
$\displaystyle =\frac{8}{6}(6-2x)$
which equals zero when x=3
$\displaystyle x=3,\ y=\frac{8}{6}(6-3)=4$