Applied Calculus - Please verify my answer?

Question

Two banks offer different investment opportunities. Bank A offers 8.75% compounded daily, and bank B offers 8.5% compounded continuously. Which is the better investment?Support your answer with appropriate details.

This is the answer I came up with:

Assume:
P=$1
t=1 year

compounded daily:
A=1(1+.0875/365)^365*1
A=$1.09

compounded continuously:
e=lim n=>infinity (1+(1/n))^n

therefore Bank a offers the better return on the investment in the long term, because as n gets larger without bound, A approaches e.

Am I even close? Appreciate your input.

Answer 1

Bank A is better

I can't remember the math involved, but the formula for continuously compounded interest is A=Pe^(rt)
where e is the natural exponent.

In this case, A=1*e^(.085*1) which is 1.08872 and bank A was 1.09143

Okay, e is the limit as n=> infinity but more specifically, it is e^1 because your expression was (1+(1/n))^n.
If you make it (1+r/n)^n then the limit becomes e^r.
When you add the time to the expression (1+r/n)^(nt) you get e^(rt).

Answer 2

Banks generally assume all months have 30 days for the purpose of calculating interest when it is compounded daily. Let's assume we start with amount $1000. Calculate the amount P after 1 year.

For bank A we have:

P = 1000(1 + 0.0875/360)^360 = $1,091.43

For bank B we have:

P = 1000e^(0.085*1) = $1,088.72

So bank A is the better deal.

Answer 3

see first evaluate the dimensions of the cable in terms of x.it incredibly is going to be 2 circumstances the dimensions of the hypotenuse from (x,0) till ultimately (9,0)+length of the cord from (x,0) uptill (12.0) symbolically it would be F(x)=2(?9^2+x^2) + (12-x) Now diffrentiate the function and equate it to 0.discover x. exchange this fee of x into the function to get the minimum lenght of the cord required to place the cables.