Analysis question

• February 21st 2009, 03:37 PM
scubasteve94
Analysis question
A rectangle ABCO has two vertices, C and B, on the x-axis and y-axis respectively and another vertex, A, on the graph of 2x+y=4, as shown in the diagram. the coordinates of A is (a,b) where a and b are positive real numbers.

i) find the area, R, of rectangle ABCO in terms of a.

ii) What is the possible value(s) of a, so that the rectangle ABCO exists?

iii) Sketch the graph of R.

iv) Find the maximum value of R and the value of a for which this occurs.

IF I WERE TO DO IT I WOULD:
i) R = 2aX2b

ii) possible value(s) of a = ≤1

iii) Unsure how to this part? (Doh)

iv) i think it has to do with previous question so unsure again (Doh)(Worried)
• February 21st 2009, 04:39 PM
skeeter
Quote:

Originally Posted by scubasteve94
A rectangle ABCO has two vertices, C and B, on the x-axis and y-axis respectively and another vertex, A, on the graph of 2x+y=4, as shown in the diagram. the coordinates of A is (a,b) where a and b are positive real numbers.

i) find the area, R, of rectangle ABCO in terms of a.

ii) What is the possible value(s) of a, so that the rectangle ABCO exists?

iii) Sketch the graph of R.

iv) Find the maximum value of R and the value of a for which this occurs.

IF I WERE TO DO IT I WOULD:
i) R = 2aX2b

$R = a(4 - 2a)$

ii) possible value(s) of a = ≤1

$0 < a < 2$

iii) Unsure how to this part?

$R = x(4-2x)$ ... graph is an inverted parabola

iv) i think it has to do with previous question so unsure again

maximum value occurs at the vertex of the parabola

.
• February 21st 2009, 04:58 PM
Soroban
Hello, scubasteve94!

Quote:

A rectangle $ABCO$ has two vertices, $C, B$ on the x-axis and y-axis, respectively.
. . and another vertex $A$ on the line $2x+y\:=\:4$.
The coordinates of $A$ are $(a,b)$ where $a, b > 0.$

Code:

    \|     4*       |\       | \       |  \       |  \       |    \  A     B * - - * (a,b)       |    |\       |    b| \       |    |  \   - - + - - * - * - -       O  a  C  2\
Quote:

i) find the area, $R$, of rectangle $ABCO$ in terms of $a.$

The line has the equation: . $y \:=\:4-2x$

If $x = a$, then: . $b \:=\:4-2a$

The area of the rectangle is: . $a\cdot b \:=\:a(4-2a)$

Therefore: . $R \;=\;4a - 2a^2$

Quote:

ii) What is the possible value(s) of $a$, so that the rectangle $ABCO$ exists?

On the graph, we see that $a$ can range from 0 to 2.

Therefore: . $0 \:< \:a \:<\:2$

Quote:

iii) Sketch the graph of $R.$
The graph of: $R \:=\:4a-2a^2$ is a parabola.
It opens downward and has x-intercepts at 0 and 2.

The graph looks like this:
Code:

      R|         |      *         |  *      *         | *          *         |*            *         |     - - * - - - - - - - * - - a         0              2

Quote:

iv) Find the maximum value of $R$ and the value of $a$ for which this occurs.

The maximum value of the parabola occurs at its vertex.

The vertex is at $a = 1$

. . and: . $R \:=\:4(1) - 2(1^2) \:=\:2$

• February 21st 2009, 07:16 PM
scubasteve94
SOROBAN

Thank you so much
it was very helpful and i now fully understand the question
thanks again
(Happy)(Rofl)(Cool)