Preclac semester review - A little urgent I guess

[anewera]

Thanks for looking at the thread and thanks in advance for taking your time to help me decimate the semester exam .Well, the semester exam is in 3 days and I got through the huge study guide with no problems...well, almost. There's 1 question I'm missing a tiny little link on, and another question that I have no idea how to start.

1. At the beginning of each week a scientist adds 200 bacteria to a population that grows at a rate of 4% weekly. If they do a 10 week experiment from beginning to end, how many bacteria are there in the end?

My attempt: Well I set this up using this equation A[sub:152ktsza]0[/sub:152ktsza] = 200(1+0.04)[sup:152ktsza]10[/sup:152ktsza] and then I noticed that he's also adding 200 bacteria EVERY WEEK so the 4% amount changes every time. I tried changing the rate from 1.04 to 2.04 but it doesn't double every week so that was obviously off. I just can't remember the method but I'm sure I'm not getting the right answer either way no matter how sensible it may seem at first glance.

2. An equilateral triangle is formed so that one of its sides is on the x-axis and such that the top vertex is on the curve defined by f(x) = -2x[sup:152ktsza]3[/sup:152ktsza] + 10x[sup:152ktsza]2[/sup:152ktsza] - x + 5 in the first quadrant. Write the area (A) of the triangle in terms of x. What is the domain of A(x)? What is the maximum area and the x value that gives the maximum?

My attempt: Well for the first part I'm not even quite sure where to start. All the sides must be equal, yes, but I don't know what to do with that information. I'm assuming I need to find the intersection point of the triangle and the cubic for the first and second parts of the question. The maximum area and the x value I can find by graphing the answer from part A onto the nSpire. ^^

I'm hoping for an answer today (It's 9:41 here in Japan right now) but as long as it's before Monday, it'll be great.

 

[soroban]

Hello, anewera!

1. At the beginning of each week a scientist adds 200 bacteria
to a population that grows at a rate of 4% weekly.
If they do a 10-week experiment, how many bacteria are there in the end?

If this had been a compound investment problem, your approach was correct.
. . "$200 is deposited and earns 4% interest per year.
. . How much is in the account at the end of 10 years?"
The answer would be: .\(\displaystyle A \;=\;200(1+0.04)^{10}\)

But this problem is an annuity, where periodic deposits are made.
. . "$200 is deposited annually and earns 4% interest.
. . How much is in the account at the end of 10 years?"
This is a formula for this problem, but we can derive it ourselves.


\(\displaystyle \text{Year 1: \$200 is deposited and earns 4\% interest.}\)
\(\displaystyle \text{Balance: }\:A_1 \;=\;200(1.04)\)

\(\displaystyle \text{Year 2: \$200 is deposited and earns 4\% interest.}\)
\(\displaystyle \text{Balance: }\:A_2 \;=\;\bigg[200(1.04) + 200\bigg](1.04) \;=\;200\bigg[1.04^2 + 1.04\bigg]\)

\(\displaystyle \text{Year 3: \$200 is deposited and earns 4\% interest.}\)
\(\displaystyle \text{Balance: }\:A_3\;=\;\bigg[200(1.04^2 + 1.04) + 200\bigg](1.04) \;=\;200\bigg[1.04^3 + 1.04^2 + 1.04 + 1\bigg]\)

\(\displaystyle \text{We can see the pattern:}\)
. . \(\displaystyle A_n \;=\;200\bigg[1.04^n + 1.04^{n-1} + \hdots + 1.04 + 1\bigg]\)


\(\displaystyle \text{We have: }\:A_n \;=\;200\underbrace{\bigg[1 + 1.04 + 1.04^2 + \hdots + 1.04^n\bigg]}_{\text{geometric series}}\)

\(\displaystyle \text{The geometric series has: first term }a=1\text{, common ratio }r = 1.04\text{, and }n+1\text{ terms.}\)

. . \(\displaystyle \text{Its sum is: }\:S_n \;=\;\frac{1.04^n - 1}{1.04-1} \;=\;\frac{1.04^n-1}{0.04}\)

\(\displaystyle \text{Hence: }\:A_n \;=\;200\,\frac{1.04^n-1}{0.04}\)


\(\displaystyle \text{Therefore: }\;A_{10} \;=\;200\,\frac{1.04^{10}-1}{0.04} \;\approx\; 2401\text{ baceteria}\)

 

[anewera]

Wow, thanks so much for the great explanation. Exactly what I needed. Thanks. One down, one to go ^^