Problem involving maxima & minima (1 Viewer)

[Smeegen999]

The cost of runnning a car at an average spped of v km/h is given by:

c = 150 + [v^2 / 80] cents per hour.


Find the average speed, to the nearest km/h, at which the cost of a 500km trip is a minimum.

Help please! :angry:

 

[undalay]

The cost of runnning a car at an average spped of v km/h is given by:

c = 150 + v squared cents per hour.
80

Find the average speed, to the nearest km/h, at which the cost of a 500km trip is a minimum.

Help please! :angry:

The time taken for the 500km journey is . 500/v (in hours)
the cost is 150+v^2 for every hour.

thus overall cost is 500/v x (150+v^2)

 

[tommykins]

I don't understand what the 80 is doing there, can you please clarify?

 

[Smeegen999]

I suggest denoting your fractions in the following manner


(a/b)

[a/(b+c)]

This will make it easier for people to understand.

ok i've edited my post

 

[SpinCobra]

Holy crap. I forgot all about Maximum and minimum.

I have no idea how to do this anymore. Damnit. Whew, glad I found this thread. =D

Do you have the answer to the question? Is it 77km/h? I have a weird feeling I'm way off..

Nevermind. I'm horribly wrong.

 

[SpinCobra]

Okay I got it. I think.

C = (500/v)(150 + v²/80)

= 75000v^-1 + 25v/4

C' = -75000v^-2 + 25/4

St. Points at C' = 0

75000v^-2 = 25/4

300000v^-2 = 25

1/v² = 25/300000

v² = 300000/25

v = 110, -110

v >= 0

Therefore 110km/h

 

[HeavenlyAnarchy]

I am doing 14 units, most are going well except maths. One of my pitfalls is these minima and maxima problems. I seem to not understand the real life application and how to rewrite the equations.

They just seem like jumble. XD