# Thread: pre-calculus urgent help with functions

1. ## pre-calculus urgent help with functions

A rectangle is created with two left hand vertices on the y-axis. One of the other vertices is on the line y=10-x and the other is on the line y=2x-5. Assuming the rectangle is bounded by the lines and the y-axis.
1) write a formula for the area of the rectangle
2) what is the domain of A(x)?
3) find the area when x=2
4) find the maximum area of the rectangle
PLEASE EXPLAIN HOW YOU FOUND ALL 4 VERTICES
Please show your work in detail, thanks for taking time to do this!!!

2. Let the right hand side of the rectangle be x = k. (Left hand side is x = 0).
Right hand vertices are: (k, y1) and (k, y2) where
y1 = 10 - k and y2 = 2k - 5.
1. Area A(k) = (y1 - y2) (k - 0) = 5k(3-k)
2. Domain of A(k) is bounded by k >= 0 and 3 - k >= 0
3. Substitute k=2
4. What is the max of k(3-k) as k goes from 0 to 3?
(If you don't know derivatives, think of area of a rectangle that has a constant perimeter of 2k + 2(3-k) = 6 units. When is area maximum?)

3. Hello, shirleychen888!

Did you make a sketch?

A rectangle is created with two left hand vertices on the y-axis.
One of the other vertices is on the line $y\:=\:10-x$
. . and the other is on the line $y\:=\:2x-5$
Assuming the rectangle is bounded by the lines and the y-axis:

1) Write a formula for the area of the rectangle
Code:
      | y = 10-x
10 *           *
| *        *
|   *     *
+-----*  *
|:::::| *
|:::::|*  *
+-----*     *
|    *        *
- - + - * - - - - - * - - -
|  *           10 *
| *                 *
|* y = 2x-5
-5 *
|
The area of a rectangle is: . $\text{(Area)} \:=\: \text{(Base)} \times \text{(Height)}$

Our rectangle has: . $\text{Base }= x$
and: . $\text{Height} \:=\:(10-x) - (2x-5) \:=\:15 - 3x$

Therefore: . $A(x) \;=\;x(15-3x)$

2) What is the domain of $A(x)$ ?
Domain: . $0 \:\leq\: x \:<\:5$

3) Find the area when $x=2$
$A(2) \;=\;2(15 - 3\cdot 2) \;=\;2\cdot9 \;=\;18$

4) Find the maximum area of the rectangle
The graph of: $A \:=\:15x-3x^2$ is a down-opening parabola.
. . Its maximum is at its vertex.

The vertex is at: . $x \:=\:\frac{\text{-}15}{\text{-}6} \:=\:\frac{5}{2}$

The maximum area is: . $A\left(\frac{5}{2}\right) \:=\:15\left(\frac{5}{2}\right) -3\left(\frac{5}{2}\right)^2 \;=\;\frac{75}{4}$