A compound produced in the laboratory has a growth equation given by $w = \frac{600}{10 + 30e^{-0.5t}}$ where $w$ is the mass in grams and $t$ is the time in hours after the compound was first produced. Find

1. Substitute $t = 0$ into the equation for $w$

When $t = 0$,
$w = \frac{600}{10 + 30e^{-0.5(0)}}$

2. Evaluate the expression to find the initial mass

$w = \frac{600}{10 + 30(1)}$
$w = \frac{600}{40}$
$w = 15$

1. Convert the given time into hours

$t = 5 + \frac{15}{60} = 5.25\text{ hours}$

2. Substitute the time into the growth equation to find the mass

$w = \frac{600}{10 + 30e^{-0.5(5.25)}}$
$w = \frac{600}{10 + 30e^{-2.625}}$
$w \approx 49.2886$
$w \approx 49.3\text{ g}$ (to 3 s.f.)

1. Substitute the given mass into the equation

$30 = \frac{600}{10 + 30e^{-0.5t}}$

2. Rearrange the equation to isolate the exponential term

$10 + 30e^{-0.5t} = \frac{600}{30}$
$10 + 30e^{-0.5t} = 20$
$30e^{-0.5t} = 10$
$e^{-0.5t} = \frac{1}{3}$

3. Solve for $t$ using natural logarithms

$-0.5t = \ln\left(\frac{1}{3}\right)$
$-0.5t = -1.0986$
$t = 2.1972$
$t \approx 2.20 \text{ hours (to 3 s.f.)}$

1. Determine the behaviour of the exponential term as time increases indefinitely

As $t \to \infty$, $e^{-0.5t} \to 0$.

2. Calculate the limiting mass and conclude

Therefore, $w \to \frac{600}{10 + 30(0)} = \frac{600}{10} = 60$.
The mass will stabilise to a final mass of 60g.