Hello, scubasteve94!
A rectangle $\displaystyle ABCO$ has two vertices, $\displaystyle C, B$ on the xaxis and yaxis, respectively.
. . and another vertex $\displaystyle A$ on the line $\displaystyle 2x+y\:=\:4$.
The coordinates of $\displaystyle A$ are $\displaystyle (a,b)$ where $\displaystyle a, b > 0.$ Code:
\
4*
\
 \
 \
 \
 \ A
B *   * (a,b)
 \
 b \
  \
  +   *  *  
O a C 2\
i) find the area, $\displaystyle R$, of rectangle $\displaystyle ABCO$ in terms of $\displaystyle a.$
The line has the equation: .$\displaystyle y \:=\:42x$
If $\displaystyle x = a$, then: .$\displaystyle b \:=\:42a$
The area of the rectangle is: .$\displaystyle a\cdot b \:=\:a(42a)$
Therefore: .$\displaystyle R \;=\;4a  2a^2$
ii) What is the possible value(s) of $\displaystyle a$, so that the rectangle $\displaystyle ABCO$ exists?
On the graph, we see that $\displaystyle a$ can range from 0 to 2.
Therefore: .$\displaystyle 0 \:< \:a \:<\:2$
iii) Sketch the graph of $\displaystyle R.$ The graph of: $\displaystyle R \:=\:4a2a^2$ is a parabola.
It opens downward and has xintercepts at 0 and 2.
The graph looks like this: Code:
R
 *
 * *
 * *
* *

  *        *   a
0 2
iv) Find the maximum value of $\displaystyle R$ and the value of $\displaystyle a$ for which this occurs.
The maximum value of the parabola occurs at its vertex.
The vertex is at $\displaystyle a = 1$
. . and: .$\displaystyle R \:=\:4(1)  2(1^2) \:=\:2$