As the financial consultant to a classic auto dealership, you estimate that the total value (in dollars) of its collection of 1959 Chevrolets and Fords is given by the formula
v = 306,000 + 1,000t2 (t ≥ 5)
where t is the number of years from now. You anticipate a continuous inflation rate of 5% per year, so that the discounted (present) value of an item that will be worth $v in t years' time is
p = ve−0.05t.
When would you advise the dealership to sell the vehicles to maximize their discounted value? (Round your answer to one decimal place.)
ugugugu
To maximize the discounted value of the collection, we need to find the time when the discounted value is highest.
The discounted value of an item that will be worth $v in t years' time is given by the formula:
p = ve^(-0.05t)
In this case, v is given by the formula:
v = 306,000 + 1,000t^2 (t ≥ 5)
We want to find the value of t that maximizes p. To do this, we can differentiate p with respect to t and set it to zero.
Let's differentiate p with respect to t:
dp/dt = d/dt (v * e^(-0.05t))
= dv/dt * e^(-0.05t) + v * d/dt(e^(-0.05t))
To find the maximum, we need dp/dt = 0, so we set the equation to zero:
0 = dv/dt * e^(-0.05t) + v * d/dt(e^(-0.05t))
Now, let's plug in the expression for v:
0 = d/dt (306,000 + 1,000t^2) * e^(-0.05t) + (306,000 + 1,000t^2) * d/dt(e^(-0.05t))
We can differentiate the terms on the left side of the equation:
0 = (2,000t * e^(-0.05t)) + (306,000 + 1,000t^2) * (-0.05 * e^(-0.05t))
Simplifying the equation further:
0 = 2,000t * e^(-0.05t) - 0.05(306,000 + 1,000t^2) * e^(-0.05t)
Now, we can factor out e^(-0.05t):
0 = e^(-0.05t) * (2,000t - 0.05(306,000 + 1,000t^2))
Since e^(-0.05t) is always positive, we can set the second factor equal to zero:
2,000t - 0.05(306,000 + 1,000t^2) = 0
Let's solve this equation for t:
2,000t = 0.05(306,000 + 1,000t^2)
40t = 306,000 + 1,000t^2
1,000t^2 - 40t + 306,000 = 0
Now we have a quadratic equation. We can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1,000, b = -40, and c = 306,000.
t = (-(-40) ± √((-40)^2 - 4(1,000)(306,000))) / (2(1,000))
t = (40 ± √(1,600 + 1,224,000)) / 2,000
t = (40 ± √1,225,600) / 2,000
t = (40 ± 1,105.65) / 2,000
Now we have two possible values for t:
t1 = (40 + 1,105.65) / 2,000
t2 = (40 - 1,105.65) / 2,000
Considering the requirement that t ≥ 5, we can see that t2 is not a valid solution.
Therefore, the advised time to sell the vehicles to maximize their discounted value is approximately:
t = (40 + 1,105.65) / 2,000
t ≈ 0.5733 years
Rounding to one decimal place:
t ≈ 0.6 years
So, the recommended time to sell the vehicles to maximize their discounted value is approximately 0.6 years from now.