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Topic review (newest first)

bob bundy
2012-03-26 02:43:28

hi caymanisland and Naru

Here's what I'm getting:

running cost (v) = 15(0.9 + 0.0016v^2) for this journey (15 lots of 100)

driver cost = (1500/v) x 30  (journey time x cost per hour)

so



so differentiate



Set  this equal to zero and re-arrange





This is > 0 for the calculated v so this gives a minimum.

Bob

bobbym
2012-03-25 19:37:44

Hi Bob;

The only way I can see to get the answer they want is to treat her salary of $30 per hour as part of the cost. The question stating she pays herself suggests that.



We differentiate and set to 0.



Solving we get one real root of v =97.87 km / hr. That is the minimum velocity

bob bundy
2012-03-25 18:54:42

hi bobbym,

Your post just got in ahead of mine.  Maybe a different interpretation?  Post 6

Bob

bobbym
2012-03-25 18:48:38

Hi Naru;

What is the name of the text book?

bob bundy
2012-03-25 18:47:59

hi caymanisland and Naru

Welcome, both of you, to the forum.  smile

I'm interpreting this problem differently.

She has two costs when she makes trips.

(i) The running costs go up as the velocity of the truck increases.

(ii) But she is making a charge for her time (to the customers?) which goes up with the time.

So the overall cost goes higher with speed due to runnning costs but lower due to her time costs.

Anyone want to try that? 

Bob

Naru
2012-03-25 11:33:48

gAr wrote:

Hi caymanisland,

The cost of the trip is an increasing function, so I assume that you mean that the cost must not increase beyond 30 $/hr.
I solved the following equation, which has one real solution:


x is the speed in km/hr, so that cost is 30 $/hr. Any speed less than that would result in lower $/hr as well as lower cost of the trip.

The answer in my textbook says it's 97.9km/h. Are you sure that's right?

gAr
2011-03-29 19:23:42

And it costs even when it's at rest.

bobbym
2011-03-29 18:59:59

Hi caymanisland;

Welcomes to the forum. Are you sure you copied the problem correctly? You should copy them word for word.

If we set up the equation of Brenda's cost and following gAr's idea.



v = 121.7901312369059 km/hr

But like him I am forced to make some assumptions. With this model there is no minimum.

gAr
2011-03-29 16:29:04

Hi caymanisland,

The cost of the trip is an increasing function, so I assume that you mean that the cost must not increase beyond 30 $/hr.
I solved the following equation, which has one real solution:


x is the speed in km/hr, so that cost is 30 $/hr. Any speed less than that would result in lower $/hr as well as lower cost of the trip.

caymanisland
2011-03-29 13:17:18

Brenda drives an 18 wheeler.   she plans to buy her own truck. Her research indicates that the expected running costs, C, in dollars, per 100 km, are given by C(v)=0.9 +0.0016v square, where v is the speed, in kilometers per hour.   Brenda's first trip will be 1500 km, round trip. she plans to pay herself $30/h.  determine the speed that will minimize Brend's cost for the trip

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