Suppose that the cost of operating a truck in Mexico is 54 +0.29 v cents per mile when the truck runs at a steady speed of v miles per hour. The top speed of the truck is 100 mph. Assume that the driver is paid 9 dollars per hour to drive the truck, and he is to begin a 2900 mile trip.
Write the cost of operating the truck in dollars, as a function of the speed v, for the planned trip .
Write the cost of driver's wages in dollars, as a function of the speed v, for the planned trip .
The total cost of the planned trip, as a function of the speed v, is the sum of the first two costs. Find the most economic speed for the planned trip, i.e., the speed that minimize the total cost is v=
2006-11-17 05:18:33 · 4 answers · asked by Lionheart12 5 in Science & Mathematics ➔ Mathematics
a) 2900 * (54 + 0.29 v)
b) 9 * 2900/v
c) 2900 * (54 + 0.29 v + 9/v)
v = 5.57
2006-11-17 05:26:34 · answer #1 · answered by feanor 7 · 1⤊ 0⤋
Cost(v) = Cost per mile * miles
Cost(v) = (54 + 0.29v) * 100 * 2900
Wage(v) = (Miles/MPH) * hourly_wage
Wage(v) = (2900/v) * 9
TotalCost(v) = Cost(v) + Wage(v)
TotalCost(v) = (54 + 0.29v) * 100 * 2900 + (2900/v) * 9
However this is all wrong since it doesn't cost 54 + .29v cents per mile, then it would cost $5429 a mile to operate the truck. I'm going to assume that you meant 54 + 0.29v dollars per mile and change it to:
TotalCost(v) = (54 + 0.29v) * 2900 + (2900/v) * 9
Then we need to find the minimum of this. Take the derivative, solve for zero and we get 5.57 or so with bounds of 0-100. Plug this in and see that the total cost is less here than any speed before or after it so it must be a minimum, thus we have the most cost effective speed.
2006-11-17 05:35:30 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Total cost: (54 + 0.29v) * 2900 + 9 * 2900 / v = C + 2900 * (0.29 v + 9 / v)
Minimize 0.29 v + 9 / v subject to 0 < v <= 100. You can do this by taking the derivative and setting it equal to zero and solving. Take the points you get, discarding any above 100. Check the cost at each of these points and choose the one with the lowest cost. If that cost is lower than the cost at v = 100, report that speed; otherwise, report 100.
2006-11-17 05:29:37 · answer #3 · answered by Charles G 4 · 1⤊ 0⤋
enable's say the oblong area of the room is 2x = W huge and L lengthy. Then the element of the rectangle area is 2xL and the element of the semicircle area is two * a million/2 * x^2 * pi = snap shots^2 So the section is 2xL + snap shots^2. The rectangle area of the circumference is 2L. The semicircle area of the circumference is two * a million/2 * 2xpi = 2xpi So the circumference is 2L + 2xpi = 2 hundred So L + xpi = one hundred So L = one hundred - xpi Substituting contained in the unique section, we get section = 2xL + snap shots^2. section = 2x(one hundred - xpi) + snap shots^2 section = 200x - 2pix^2 + snap shots^2 section = 200x - snap shots^2 section' = 2 hundred - 2pix 2 hundred - 2pix = 0 2pix = 2 hundred snap shots = one hundred x = one hundred/pi = 31.80 3 L = one hundred - xpi L = one hundred - 31.80 3 * pi the project that I have right that is that L = 0. Which in a way is smart. a thanks to maximise section for a given circumference is to make a circle. even as that's a circle, section = 200x - snap shots^2 section = 2 hundred * 31.80 3 - pi * (31.80 3)^2 section = 3183.10
2016-11-25 00:46:29 · answer #4 · answered by ? 3 · 0⤊ 0⤋