# this is applied maximum and minimum problems

• Mar 3rd 2010, 11:18 PM
redcherry
this is applied maximum and minimum problems
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..(Thinking)
tq~
• Mar 4th 2010, 03:41 AM
Quote:

Originally Posted by redcherry
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..(Thinking)
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

$tan\theta=\frac{y}{6-x}=\frac{8}{6}\ \Rightarrow\ 6y=8(6-x)\ \Rightarrow\ y=\frac{8}{6}(6-x)$

Rectangle area is

$xy=\frac{8x}{6}(6-x)$

in terms of x only, allowing us to differentiate the area and equate the derivative to zero to find the rectangle of maximum area

$\frac{8}{6}\frac{d}{dx}\left[x(6-x)\right]=\frac{8}{6}\frac{d}{dx}\left(6x-x^2\right)$

$=\frac{8}{6}(6-2x)$

which equals zero when x=3

$x=3,\ y=\frac{8}{6}(6-3)=4$