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~

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- Mar 3rd 2010, 10:18 PMredcherrythis 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, 02:41 AMArchie Meade
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

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