1. ## Optimization problem?????

a rectangle is to be inscribed in a right triangle having sides 6 inches, 8 inches, and 10 inches. Determine the dimensions of the rectangle with greatest area.

I recently tried this question and the answer was found by finding the slope and then using the first and second derivatives of the area but it turns out that I was wrong.
Now I would like to know how to solve this by using proportions and the derivative tests. So if there are any suggestions whatsoever, thank you because I really don't understand it.

2. draw a set of coordinate axes ... also, sketch the line $\displaystyle y = -\frac{4}{3}x + 8$

note that the line forms a 6-8-10 right triangle in quadrant I.

inscribe a rectangle in the triangle with sides parallel to the coordinate axes.

area of the rectangle is ...

$\displaystyle A = xy$

$\displaystyle A = x\left(8 - \frac{4}{3}x\right)$

find $\displaystyle \frac{dA}{dx}$ and maximize.

3. 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.

4. Originally Posted by skeeter
draw a set of coordinate axes ... also, sketch the line $\displaystyle y = -\frac{4}{3}x + 8$

note that the line forms a 6-8-10 right triangle in quadrant I.

inscribe a rectangle in the triangle with sides parallel to the coordinate axes.

area of the rectangle is ...

$\displaystyle A = xy$

$\displaystyle A = x\left(8 - \frac{4}{3}x\right)$

find $\displaystyle \frac{dA}{dx}$ and maximize.
I already did it this way and thats not how I am supposed to find the answer. its wrong for the way I am supposed to solve it.

5. ... then how, pray tell, are you supposed to solve it?

6. Originally Posted by skeeter
... then how, pray tell, are you supposed to solve it?
My professor is saying that I am supposed to use proportions with the x and y sides like subtracting the areas of the smaller triangles that are made when the rectangle is inscribed in the right triangle.
do you understand?

7. yes, I do understand.

use your sketch ... let the large triangle base = 8, height = 6.

area of rectangle = 24 - area of two smaller similar triangles

$\displaystyle A = 24 - \frac{1}{2}xy - \frac{1}{2}(8-x)(6-y)$

$\displaystyle \frac{y}{x} = \frac{6}{8} = \frac{3}{4}$

$\displaystyle 3x = 4y$

$\displaystyle x = \frac{4y}{3}$

$\displaystyle A = 24 - \frac{1}{2}\left(\frac{4y}{3}\right)y - \frac{1}{2}\left(8-\frac{4y}{3}\right)(6-y)$

$\displaystyle A = 24 - \frac{2y^2}{3} - \left(\frac{2y^2}{3} - 8y + 24\right)$

$\displaystyle A = 8y - \frac{4y^2}{3}$

now find $\displaystyle \frac{dA}{dy}$ and maximize ... you should get y = 3 , then x = 4.

hope your prof is satisfied with his "back door" method.

8. Exactly. I knew someone would figure it out.
Thanks for showing the steps. I hope my pro. appreciates this.