Calculus Help. Test next period

[MrCodeDude]

The cost of fuel for running a locomotive is proportional to the 3/2 power of the speed (s^(3/2) and is $50 per hour for a speed of 25 miles per hour. Other fixed costs amount to an average of $100 per hour. Find the speed that will minimize the cost per mile.

I have the cost equation as:

c = 3/2 * s^(3/2) + $150 h

$50 h = 25s
h = 1/2 s

Probably wrong, but I have like 20 minutes until the test 🙁

 

[MrCodeDude]

Bump. Please help.

 

[MrCodeDude]

🙁

 

[TTM77]

This I believe is more physics.

I don't believe "proportional" is ^ (power of). I think it's more 2S/3.

So.. it's $50C/25 = 2S/3

C * 50/25 = 2S/3

C =25/50 * 2S/3

C = S/3

 

[MrCodeDude]

I am so screwed.

 

[crt1530]

🙂

 

[TTM77]

😀

 

[Oscar1613]

that question is worded so weird i dont know what the hell they're trying to say... otherwise i'd try to help😕

 

[MrScott81]

Originally posted by: Oscar1613
that question is worded so weird i dont know what the hell they're trying to say... otherwise i'd try to help😕

 

[dighn]

cfuel = ks^3/2, 50 = k 25^3/2 => k=2/5

c = 2/5*s^3/2+100

cost per mile = cost/hour / mile/hour = c/s = 2/5sqrt(s)+100/s

take the derivative of that and solve for 0 for critical points, and test the end points also

 

[SpazzyChicken]

Originally posted by: dighn
cfuel = ks^3/2, 50 = k 25^3/2 => k=2/5

c = 2/5*s^3/2+100

cost per mile = cost/hour / mile/hour = c/s = 2/5sqrt(s)+100/s

take the derivative of that and solve for 0 for critical points, and test the end points also

Bingo! Now, if you can't take the derivative and solve for the critical points, you really are screwed!

 

[MrCodeDude]

Originally posted by: dighn
cfuel = ks^3/2, 50 = k 25^3/2 => k=2/5

c = 2/5*s^3/2+100

cost per mile = cost/hour / mile/hour = c/s = 2/5sqrt(s)+100/s

take the derivative of that and solve for 0 for critical points, and test the end points also

Thanks for the help. Too bad this question or a question similar to this question was not on the test 🙁