Given the function $f(t) = 18 + 3t - t^2$ representing the amount of money in the bank (in thousand dollars) after $t$ years, we can analyze and answer each part of the problem.

Part (a): Increase During Year 3

To find the increase during year 3, we need to evaluate the function at $t = 3$ and $t = 2$:

$f(3) = 18 + 3(3) - (3)^2 = 18 + 9 - 9 = 18$ $f(2) = 18 + 3(2) - (2)^2 = 18 + 6 - 4 = 20$

The increase during year 3 is:

$\Delta f = f(3) - f(2) = 18 - 20 = -2$

So, the bank account decreased by 2 thousand dollars during year 3.

Part (b): Average Rate of Increase During Year 3

The average rate of increase is given by the change in the function value over the change in time:

$\text{Average Rate} = \frac{f(3) - f(2)}{3 - 2} = \frac{18 - 20}{1} = -2 \text{ thousand dollars per year}$

Part (c): Instantaneous Rate of Increase at the Start of Each Quarter

To find the instantaneous rate of increase, we need the derivative $f'(t)$:

$f(t) = 18 + 3t - t^2$ $f'(t) = 3 - 2t$

Now, evaluate the derivative at the start of each quarter of year 3:

• Start of first quarter: $t = 2$
• Start of second quarter: $t = 2.25$
• Start of third quarter: $t = 2.5$
• Start of fourth quarter: $t = 2.75$

c1) Start of First Quarter (t = 2)

$f'(2) = 3 - 2(2) = 3 - 4 = -1 \text{ thousand dollars per year}$

c2) Start of Second Quarter (t = 2.25)

$f'(2.25) = 3 - 2(2.25) = 3 - 4.5 = -1.5 \text{ thousand dollars per year}$

c3) Start of Third Quarter (t = 2.5)

$f'(2.5) = 3 - 2(2.5) = 3 - 5 = -2 \text{ thousand dollars per year}$

c4) Start of Fourth Quarter (t = 2.75)

$f'(2.75) = 3 - 2(2.75) = 3 - 5.5 = -2.5 \text{ thousand dollars per year}$

Part (d): When Does the Bank Account Run Out of Money?

The bank account runs out of money when $f(t) = 0$:

$18 + 3t - t^2 = 0$

Rearrange to form a quadratic equation:

$t^2 - 3t - 18 = 0$

Solve using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$t = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-18)}}{2(1)}$ $t = \frac{3 \pm \sqrt{9 + 72}}{2}$ $t = \frac{3 \pm \sqrt{81}}{2}$ $t = \frac{3 \pm 9}{2}$

So, we have two solutions:

$t = \frac{12}{2} = 6$ $t = \frac{-6}{2} = -3$

Since $t$ represents time in years, we discard the negative solution. Therefore, the bank account runs out of money after 6 years.

If you need further details or have any questions, please let me know. Here are some questions for you to consider next:

1. How can the derivative of a function help in understanding the behavior of the function?
2. What is the significance of the quadratic formula in solving polynomial equations?
3. How do we interpret the instantaneous rate of change in real-life scenarios?
4. What are the practical implications of a negative average rate of change?
5. How can we apply these mathematical concepts to other fields, such as economics or physics?

Tip: When solving quadratic equations, always check the discriminant (under the square root) first to determine the nature of the roots (real or complex).