Norman windows consist of a rectangle topped by a semi-circle as shown above - Leaving Cert Mathematics - Question 9 - 2019

From the equation derived in (a)(i), substituting $P = 12$ gives:

$12 = 2y + (2 + \pi)x$

Rearranging gives:

$2y = 12 - (2 + \pi)x$

Dividing by 2:

$y = \frac{12 - (2 + \pi)x}{2}$

This is valid for $0 \leq x \leq \frac{12}{2 + \pi}$.

To plot the graph for the function $y = \frac{12 - (2 + \pi)x}{2}$, use the calculated values for $x$ and $y$ in the table. Plot the points (0,6) and (12,0), ensuring to connect them with a straight line.

The slope can be found from the formula:

$m = \frac{y_2 - y_1}{x_2 - x_1}$ Picking points (0, 6) and (12, 0):

$m = \frac{0 - 6}{12 - 0} = \frac{-6}{12} = -0.5$ Thus the slope is -0.50.

Interpretation: For each 1 metre increase in the radius of the semi-circle, the height of the rectangle decreases by $0.50$ m.

The area $A$ of the window can be articulated as:

$a(x) = \frac{1}{2}[2x(\text{height})] + \pi x^2$

Substituting for height:

$a(x) = \frac{1}{2}[2x \cdot \frac{12 - (2 + \pi)x}{2}] + \pi x^2$ This simplifies to: $a(x) = \frac{1}{2}(12x - (2 + \pi)x^2)$ Which resolves to: $a(x) = \frac{1}{2}[24x - (\pi + 4)x^2]$

To find the derivative of the area function, we have:

$a'(x) = \frac{1}{2}[24 - (\pi + 4)2x]\n= 12 - (\pi + 4)x$

To find the maximum area, set the derivative equal to zero:

$a'(x) = 0 \Rightarrow 12 - (\pi + 4)x = 0 \Rightarrow x = \frac{12}{\pi + 4}$

Substituting this $x$ back into the equation for $y$: $y = \frac{12 - (2 + \pi)\frac{12}{\pi + 4}}{2}$ This yields the relationship between $x$ and $y$, where the area is maximized.