Answer:
A).Amount = $218250
B). Amount = $88700
Step-by-step explanation:
A) .$5000 in an account at age 23, and withdraw it 42 years
Number of years t= 42 years
Principal P = $5000
Rate r= 9%
Number of times compounded n= 42
A= p(1+r/n)^(nt)
A= 5000(1+0.09/42)^(42*42)
A= 5000(1+0.002143)^(1764)
A= 5000(1.002143)^1764
A= 5000(43.65)
A= 218250
Amount = $218250
B).waits 10 years before making the deposit, so that it stays in the account for only 32 years
Number of years t= 32 years
Principal P = $5000
Rate r= 9%
Number of times compounded n= 32
A= p(1+r/n)^(nt)
A= 5000(1+0.09/32)^(32*32)
A= A= 5000(1+0.0028125)^(1024)
A= 5000(1.0028125)^1024
A= 5000(17.74)
A= 88700
Amount = $88700
Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of
(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min
and flows out at a rate of
(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min
Then the net flow rate is governed by the differential equation

Solve for S(t):


The left side is the derivative of a product:
![\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dt%7D%5Cleft%5Be%5E%7Bt%2F800%7DS%28t%29%5Cright%5D%3D%5Cdfrac8%7B100%7De%5E%7Bt%2F800%7D)
Integrate both sides:



There's no sugar in the water at the start, so (a) S(0) = 0, which gives

and so (b) the amount of sugar in the tank at time t is

As
, the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.