Hello, ciciasi!
A rectangle is to be inscribed in a right triangle having sides 6, 8, and 10 inches.
Determine the dimensions of the rectangle with greatest area Did you notice that we have a right triangle? Code:

(0,6)*
 *
 *
 *
+      *
 : *
 :y *
 : *
  +      +      *   
 x (8,0)
The equaton of the hypotenuse is: .$\displaystyle y \:=\:\text{}\tfrac{3}{4}x + 6$ [1]
The area of the rectangle is: .$\displaystyle A \:=\:xy$ [2]
Substitute [1] into [2]: .$\displaystyle A \;=\;x\left[\text{}\tfrac{3}{4}x+6\right] $
And we must maximize the function: .$\displaystyle A \;=\;\text{}\tfrac{3}{4}x^2 + 6x$
Got it?
Answers: .$\displaystyle x = 4,\:y = 3$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
If one side of the rectangle is on the hypotenuse,
. . we have a different (and much trickier) problem. Code:
*
* *
* * 8
6 * *
*        *
*  *
*  y *
*  x  *
* * * * * * * * *
:       10        :
Answers: .$\displaystyle x = 5,\:y = \tfrac{12}{5}$
Note: In both cases, the maximum area is 12 unitsē.