Question

what is the derivative of C(v) = (0.0004(v)^3/2 + 0.85) + 22500/(v) and what is V equal to when the derivative is set to 0?

Answer 1

So one more time. Here's your original problem:
Glynn is considering buying a truck and becoming a professinal driver. The truck manufacturer indicates that he can expect his running costs when driving at (v) kilometers per hour to be approx. c(v)=0.85+0.0004(v)^(3/2) where the cost, (c), is in dollars per kilometer. Glynn plans to pay himself 15$/h while he is driving. Find the speed that will minimize his costs for a 1500km trip

Answer 2

Given that the cost per km of the truck is c(v), for a 1500 km trip, the total cost of the truck is Capital C
C(v) = 1500 c(v)
= 1500 ( 0.85 + 0.0004v^(3/2) )
= 1275 + 0.6 v^(3/2)

The money Glynn pays himself is
15 t
where t is the time the trip takes. At an average speed of v, that time is t = 1500/v. Hence
15 t = 25000/v

Therefore the total cost of the trip, T(v), is

T(v) = C(v) + 25000/v
= 1275 + 0.6 v^(3/2) + 25000/v

Now, to minimize this function, we differentiate it.

dT/dv = 0.9 v^(1/2) - 25000/v^2

Setting that equal to zero,

v^(5/2) = 25000/0.9
= 27,777

v = 59.9 km/hr

Answer 3

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