Rates of change question

  • Thread starter ploppers
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  • #1
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Homework Statement



A person wants to move excess topsoil from his farm. He can hires a truck and driver for $60/h. The driver will take 30 mins to deliver a load of top soil. One person will take 40 h to load the truck with soil. Workers get $18/h (Including the time the truck takes). How many labourers should the person Hire?

Homework Equations



Dunno :p


The Attempt at a Solution



I figured that since it takes one person 40 hours, it should take two people 20 hours and 3 people 40/3 hours and so on...basically 40/x. That will represent the time it takes to move all the soil for one run. Then you must add the time for the driver to deliver with is 0.5 h. So the function for time is y = (40/x) + 0.5. That function will then be multiplied by the salary of both labourers and the driver which are 18x and 60. So i get:

= ((40/x) + 0.5)(18x) + ((40/x) + 0.5)(60)
= ((40/x) + 0.5)(18x + 30)
= 9x + 1200x^-1 + 735

Then I took the derivative and got
f'(x) = 9 -1200x^-2

To find the max slope should be 0

0 = 9 - 1200x^-2
1200/x^2 = 9
1200/9 = x^2

x = 11.54700

That does not match the answer in the back where it is 5 men that would make a total cost of $294
 

Answers and Replies

  • #2
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= ((40/x) + 0.5)(18x) + ((40/x) + 0.5)(60)
= ((40/x) + 0.5)(18x + 30)
= 9x + 1200x^-1 + 735
Isnt that s'posed to be [itex] \ 9x+2400x^{-1}+c [/itex]
Solving i got x=16...doesnt help :(
 
  • #3
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Anyone have a solution? Or did I make another mistake?
 
  • #4
D H
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As f(x) noted, you made a mistake here:
= ((40/x) + 0.5)(18x) + ((40/x) + 0.5)(60)
= ((40/x) + 0.5)(18x + 30)
If you have stated the problem correctly, your book has made a mistake also. It will take 8 hours for 5 workers to load the truck, making the cost for the truck and driver alone $510. The workers would have to pay you $216 for the privilege of loading the truck to make the total cost $294.

I agree with f(x): Hire 16 men.
 

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