# Math Help - maximize area: works one way, not the other

1. ## maximize area: works one way, not the other

The question is:

A right triangle represents a lot (side a=12m side b=4m). Doug wants to put a rectangular fence in that uses side a and b, and one corner touching side c of the triangle. What ar ethe dimensions that maximize the triangle.

Well, I did this question with calculus using triangle ratios.... Draw a triangle in the rectangle and use the ratio: (4-y)/4 = x/12

Now, if I solve that ratio for x and use IT in the area equation, (x)(4-y), then I get a differnt number than if I solve for y and use IT.... plus, it isn't a maximum if I use x

2. Hello, Mike!

Your wording is off . . . and your ratio is wrong . . .

A right triangle represents a lot: side a = 12m, side b = 4m.
Doug wants to put a rectangular fence in that uses side a and b,
and one corner touching side c of the triangle.
What are the dimensions that maximize the area of the rectangle?
Code:
-     *
: 4-y |  *
:     |     *
:     |        *
4     *-----------*
:     |     x     |  *
:   y |           |y    *
:     |           |        *
-     *-----------*-----------*
:     x     :   12-x    :
: - - - - - 12- - - - - :
We have two similar right triangles.

. . Hence: . $\frac{4-y}{x} \:=\:\frac{y}{12-x} \quad\Rightarrow\quad y \:=\:4 - \frac{x}{3}\;\;{\color{blue}[1]}$

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

Substitute [1]: . $A \:=\:x\left(4-\frac{x}{3}\right) \quad\Rightarrow\quad A \:=\:4x-\frac{x^2}{3}$

. . Then: . $A' \:=\:4 - \frac{2x}{3} \:=\:0 \quad\Rightarrow\quad\boxed{ x \:=\:6}$

Substitute into [1]: . $y \:=\:4-\frac{6}{3} \quad\Rightarrow\quad\boxed{ y \:=\:2}$

3. Okay, now, I did get those answers using a different ratio (on my retries)

Ratio:

(4-y)/4 = (x)/(12)

and from what I gather... I need to do these questions with 2 triangles... one triangle side must be matching one of the dimensions of the length I want, and the other one, bigger, but proportional