The weekly revenue produced by a company manufacturing air conditioning units is seasonal - Leaving Cert Mathematics - Question 8 - 2019

To find $t$, set the revenue equation equal to €26 250:

$22500 \cos \left( \frac{\pi}{26} t \right) + 37500 = 26250$

Solving for $t$ gives:

$22500 \cos \left( \frac{\pi}{26} t \right) = -11250$

This simplifies to:

$\cos \left( \frac{\pi}{26} t \right) = -0.5$

The solutions for (t) in the interval $[0, 52]$ are:

1. ( \frac{\pi}{26} t = \frac{2\pi}{3} ) gives ( t = \frac{52}{3} \approx 17.33 )
2. ( \frac{\pi}{26} t = \frac{4\pi}{3} ) gives ( t = \frac{104}{3} \approx 34.67 )

Thus, the two values of $t$ are approximately $t \approx 17.33$ and $t \approx 34.67$.

Using the chain rule for differentiation:

$r'(t) = -22 500 \sin \left( \frac{\pi}{26} t \right) \cdot \frac{\pi}{26}$

Simplifying, we have:

$r'(t) = -\frac{11250\pi}{13} \sin \left( \frac{\pi}{26} t \right)$

Calculate $r'(30)$:

$r'(30) = -\frac{11250\pi}{13} \sin \left( \frac{\pi}{26} \times 30 \right)$

First, compute ( \frac{\pi}{26} \times 30 ):

$\frac{30\pi}{26} = \frac{15\pi}{13}$

Evaluating ( \sin \left( \frac{15\pi}{13} \right) ): it is negative. Thus, $r'(30)$ becomes:

$r'(30) = -\frac{11250\pi}{13} \times (-1) > 0$

Since $r'(30) > 0$, the revenue is increasing 30 weeks after the beginning of July.