# Help with really hard optimization problem

1. Jul 25, 2011

### sniper_08

1. The problem statement, all variables and given/known data

A construction company has been offered a contract for $7.8 million to construct and operate a trucking route for five years to transport ore from a mine site to a smelter. The smelter is located on a major highway, and the mine is 3 km into a heavily forested area off the road. Construction (capital) costs are estimated as follows: •Repaving the highway will cost$200 000km.
•A new gravel road from the mine to the highway will cost $500 000km. Operating conditions are as follows: •There will be 100 round trips each day, for 300 days a year, for each of the five years the mine will be open. •Operating costs on the gravel road will be$65h, and the speed limit will be
40 kmh.
•Operating costs on the highway will be $50h, and the speed limit will be 70 kmh. Use calculus to determine if the company should accept the contract. Determine the average speeds of the trucks along the paved and gravel roads that produce optimum conditions (maximum profit). What is the maximum profit? 2. Relevant equations profit = revenue minus cost 3. The attempt at a solution i know that --> P(x) = R(x) - C(x) and R(x) = xp(x) i hate optimization and i really suck at it. i don't know how to get the equations if i get the equations then i know how to solve it. please help and thanks in advance. 2. Jul 26, 2011 ### EWH I hate it when the problem is both detailed and totally unrealistic. There are so many problems with the assumptions it's hard to know where to start. Putting all that aside, it appears that the length of highway from the smelter to the gravel road is missing, and without that the problem appears insoluble. Maybe I'm missing the point? Is there a total distance from mine to smelter provided instead? 3. Jul 26, 2011 ### sniper_08 distance of highway 10 km distance from mine to highway 3 km total distance 13 km 4. Jul 26, 2011 ### Ray Vickson The reason you suck at it is that you are not sitting down, relaxing, taking a deep breath and going at the problem step-by-small step. The first thing to do is to figure out what you want to maximize or minimize, expressed in explicit mathematical terms. Then, later, worry about how to optimize it. Suppose the junction point at x km from the smelter. You can compute the length of the sideroad from the junction to the mine, because this is the hypotenuse of a triangle with known sides (the "height" is 3 km and the "base" is related to x). So, now you know the lengths of the forest and road portions of each trip. You know the costs per km and the driving speeds on each part, so you can compute the round-trip length, time, and cost. From that, you can get the annual cost, all in terms of your unknown x. You are not given any "revenue" information, so all you can do is hope that the annual revenue from sales of the ore, etc., does not depend on x, the position of the road junction, and that seems very reasonable. So, annual cost is the only portion of annual profit that you have any control over. By the way: you need to be less careless in your writing; operating cost on the gravel road is not$65h it is $65/h, etc. Believe it or not, such simple errors can build up (in larger, more complex problems) and can lead to incorrect expressions and major blunders, etc. RGV 5. Jul 26, 2011 ### EWH Note: I could be misinterpreting this problem, but assuming it's exactly worded as you said, here's my best shot: Costs are negative, fees received are positive 10km hwy * -$2e5/km = -$2e6 3km gravel * -$5e5/km = -$1.5e6 paving total -$3.5e6
contract fee $7.8e6 fee net of paving =$4.3e6
(Assuming that the problem means that these paving jobs are a prerequisite of fulfilling the contract and that no other contingent costs for road wear depending on truck speed and/or loading are specified - which may not be realistic)

200 trips * 300 days/year * 5 years = 300,000 trips
* 10km hwy/trip = 3e6 hwy. km
*3 km gravel/trip = 9e5 gravel km

At speed limit:
-$65/hr. gravel at 40km/h = -65/40 = -$1.625/km
-$50/hr. hwy. at 70km/hr = -$5/7 /km (~-$0.7143/km) From the conditions given (if accurate) the costs per hour will remain the same at any speed, (which is not entirely realistic). The number of trips required is fixed, so no advantage comes from reducing speed and no calculus is needed - the maximum profit will be at the speed limit for each segment. So 3e6 hwy. km * ($5/7 /km) = -$(15/7)e6 = (~-$2.143e6)
9e5 gravel km * ($13/8 /km) = -$(72/13)e5 = (~$5.538e5) Total operating costs = -$2.454e8/91 = (~-$2.697e6) plus fee net of paving =$4.3e6 -$2.454e8/91 =$1.459e8/91 profit = (~$1.603e6) Average speed = d/t = (10km + 3 km)/ (10km/(70km/hr) + 3km/(40km/hr)) = 3640/61 km/hr = ~59.67km/hr *** But they want calculus, so you need an equation to optimize (with all entries below as positive) totalhwydist=(hwydist * 2 * trips/day * days/year * years) totalgraveldist=(graveldist * 2 * trips/day * days/year * years) in$:
gravelopcost[vg] = $65/hr. * totalhwydist * vg hwyopcost[vh] =$50/hr. * totalgraveldist * vh
(vg = gravel road speed, vh= highway speed)

pavingcost = (hwydist * hwypavcost + graveldist * gravelpavcost)

Profit[vg,vh] = fee - pavingcost - gravelopcost[vg] - hwyopcost[vh]
Average velocity = 13km / (10km/(vh)) + 3km/(vg))

So are you supposed to be doing multivariate calculus? Or is there some assumption as to the ratios of gravel/highway speeds? Or are the trucks supposed to travel a constant speed no matter which road type they're on?

Last edited: Jul 26, 2011