Hello, shirleychen888!
Did you make a sketch?
A rectangle is created with two left hand vertices on the yaxis.
One of the other vertices is on the line $\displaystyle y\:=\:10x$
. . and the other is on the line $\displaystyle y\:=\:2x5$
Assuming the rectangle is bounded by the lines and the yaxis:
1) Write a formula for the area of the rectangle Code:
 y = 10x
10 * *
 * *
 * *
+* *
::::: *
:::::* *
+* *
 * *
  +  *      *   
 * 10 *
 * *
* y = 2x5
5 *

The area of a rectangle is: .$\displaystyle \text{(Area)} \:=\: \text{(Base)} \times \text{(Height)} $
Our rectangle has: .$\displaystyle \text{Base }= x$
and: .$\displaystyle \text{Height} \:=\:(10x)  (2x5) \:=\:15  3x$
Therefore: .$\displaystyle A(x) \;=\;x(153x)$
2) What is the domain of $\displaystyle A(x)$ ? Domain: . $\displaystyle 0 \:\leq\: x \:<\:5$
3) Find the area when $\displaystyle x=2$ $\displaystyle A(2) \;=\;2(15  3\cdot 2) \;=\;2\cdot9 \;=\;18 $
4) Find the maximum area of the rectangle The graph of: $\displaystyle A \:=\:15x3x^2$ is a downopening parabola.
. . Its maximum is at its vertex.
The vertex is at: .$\displaystyle x \:=\:\frac{\text{}15}{\text{}6} \:=\:\frac{5}{2}$
The maximum area is: .$\displaystyle A\left(\frac{5}{2}\right) \:=\:15\left(\frac{5}{2}\right) 3\left(\frac{5}{2}\right)^2 \;=\;\frac{75}{4}$