# Logarithmic/exponential functions word problem. Easy 10 points!?

Otto owns an antique car that is presently worth \$23 000 and appreciates by 2% per month. It costs Otto \$750 per year to store his vehicle in a garage and \$350 per month to insure and maintain his car. Considering his costs, is his antique car a good investment if he keeps it for 3 years? Justify your answer with appropriate calculations.

I need to come up with an equation that will satisfy the problem, so far I know that the equation must have a "-4950(x)". Any help will be greatly appreciated. Thanks in advance.

Update:

I think that the teacher is looking for one equation and then the calculation for year three.

Relevance

Let V(t) = the value of the car after t months

We're given that the change in value with respect to time dV/dt = 0.02V. So,

dV/V = 0.02 dt

Integrating both sides, we get:

∫ dV / V = ∫ 0.02 dt

=>

ln V = 0.02t + C

=>

V(t) = Ce^(0.02t)

To find a value for the constant of integration C, we note that, at time t = 0, we're given that V(0) = \$23,000. So,

V(0) = Ce^((0.02)(0)) = 23,000

=>

C = 23,000

And, our expression for V(t) is:

V(t) = 23,000e^(0.02t)

Storage costs are given as \$750/yr = \$62.50/month. Insurance and maintenance costs are given as \$350/month. So, the total cost C(t) for storing, insuring and maintaining the car for t months is:

C(t) = 62.50t + 350t

= 412.50t

After 3 years = 36 months, the value V(36) of the car, less the cost C(36) of maintaining the car is:

V(36) - C(36)

= 23,000e^(0.02(36)) - 412.50(36)

= \$32,401.96

So, yes, the antique car is a good investment, since Otto will have made more than \$9,400 after three years.

--

Note, the reason you think the equation must have a term of -4,950x is because, if we calculate the maintenance costs in years instead of months, we get:

C(t) = 412.50t × 12

= 4,950x

However, if we did than, then we would also need to calculate the car's value V(t) in years rather than months. But, since it's easier to count t in months rather than years, we'll leave it at that.

• After 3 years (ie 36 months), the value is

23000*(1.02)^36 = 46917.40891

The value has increased by 46917.40891 - 23000 = 23917.40891 <<< PROFIT

The costs are 750*3 for storage = 2250, plus 350*36 for insurance/maintainance = 12600

Total costs 14850 <<< COSTS

Profits exceed costs by \$9067.41 to nearest cent...so it's a good investment

• car value= 23000

appreciation per year=0.24*23000=5520

costs

storage= -750

insurance etc=350*12=-4200

At the end of the year the value is 23570

after 3 years similarly calcualting we get car value as 25153.232

Hence worth keeping it