S3 2014 9 OCR

(i)

Write Y for the other side of the rectangle. We know that the perimeter of the rectangle is 20, ie. 2X + 2Y = 20, giving X + Y = 10 and so Y = 10 - X and that the area is A, so A = XY. We therefore have A = X(10 - X) = 10X - X^2. Now, completing the square:

\begin{align*}A & = 10X - X^2 \\ & = -(X^2 - 10X) \\ & = -(X^2 - 10X + 25 - 25) \\ & = 25 - (X^2 - 10X + 25) \\ & = 25 - (X - 5)^2\end{align*}

as required.

(ii)

The PDF of X is:

\displaystyle f_X(x) = \begin{cases}\frac 1 2 & 0 \le x \le 2 \\ 0 & \text{otherwise}\end{cases}

The corresponding CDF (not asked for but we’ll need it) is:

\displaystyle \mathrm F_X(x) = \begin{cases}0 & x < 0 \\ \frac 1 2 x & 0 \le x \le 2 \\ 1 & x > 2\end{cases}

(iii)

Note that since 0 \le X \le 2, we have -5 \le X - 5 \le -3 so 9 \le (X - 5)^2 \le 25. Since A = 25 - (X - 5)^2 we then have 25 - 25 = 0 \le A \le 25 - 9 = 16. The CDF of A is then given by, for 0 \le a \le 16:

\begin{align*}\mathrm F_A(a) & = \mathrm P(A \le a) \\ & = \mathrm P(25 - (X - 5)^2 \le a) \\ & = \mathrm P(25 - a \le (X - 5)^2) \\ & = \mathrm P(\sqrt {25 - a} \le |X - 5|)\end{align*}

So either \sqrt {25 - a} \le X - 5 so that 5 + \sqrt {25 - a} \le X, or X - 5 \le -\sqrt {25 - a} so that X \le 5 - \sqrt {25 - a}. Note that the former case is impossible since this would mean that X \ge 5, but 0 \le X \le 2. So we must have the latter case:

\begin{align*}\mathrm F_A(a) & = \mathrm P(\sqrt {25 - a} \le |X - 5|) \\ & = \mathrm P(X \le 5 - \sqrt {25 - a}) \\ & = \frac 1 2 \left(5 - \sqrt {25 - a}\right)\end{align*}

for 0 \le a \le 16. For a < 0 we have \mathrm F_A(a) = 0 and for a > 16 we have \mathrm F_A(a) = 1, since 0 \le A \le 16. Putting this together, we get:

\displaystyle \mathrm F_A(a) = \begin{cases}0 & a < 0 \\ \frac 1 2 \left(5 - \sqrt {25 - a}\right) & 0 \le a \le 16 \\ 1 & a > 16\end{cases}

(iv)

Differentiating, we have:

\begin{align*}\frac {\mathrm d} {\mathrm da} \left(\frac 1 2 \left(5 - \sqrt {25 - a}\right)\right) & = -\frac 1 2 \times -\frac 1 {2 \sqrt {25 - a}} \\ & = \frac 1 {4 \sqrt {25 - a}}\end{align*}

so:

\displaystyle f_A(a) = \begin{cases}\frac 1 {4 \sqrt {25 - a}} & 0 \le a \le 16 \\ 0 & \text {otherwise}\end{cases}