# Little optimization problem

#### osodud

Hello friends¡¡

I need to find the greatest area rectangle which can be drawn "inside" of a right triangle.

Measures: Cathetus a = 4cm and cathetus b = 3cm.

I need to solve the problem by using derivatives.

#### Soroban

MHF Hall of Honor
Hello, osodud!

Find the greatest area rectangle which can be drawn
inside a right triangle with legs 3 and 4.
A diagram will always help.
Code:
-   - *
:   : |  *
:  3-y|     *
:   : |        *
3   - *-----------*
:   : |     x     |  *
:   y |           |y    *
:   : |           |        *
-   - *-----------*-----------*
: - - x - - :  - 4-x -  :
: - - - - - 4 - - - - - :

The area of the rectangle is: .$$\displaystyle A \;=\;xy$$ .

From the similar right triangles, we have: .$$\displaystyle \frac{3-y}{x} \:=\:\frac{y}{4-x}$$
. . which simplifies to: .$$\displaystyle y \;=\;3-\tfrac{3}{4}x$$ .

Substitute  into : .$$\displaystyle A \;=\;x\left(3-\tfrac{3}{4}x\right)$$

And you must maximize: .$$\displaystyle A \;=\;3x - \tfrac{3}{4}x^2$$

. . Got it?