A fuel tank is being designed to contain 200 m cube of gasoline; however, the maximum length tank that can be safely transported to clients is 16 m long. The design of the tank calls for a cylindrical part in the middle with hemispheres at each end. If the hemispheres are twice as expensive per unit area as the cylindrical wall, then find the radius and height of the cylindrical part so that the cost of manufacturing the tank will be minimal. Give the answer correct to the nearest centimetre.
coldcell said:
V = 2/3 pi r³ h
200 = 2/3 pi r³ h
300 = pi r³ h
h = 300/ (pi)(r³)
So this is the relationship that I find between height and radius.
This is where I'm lost. The pricing of the cynlindrical wall and the hemisphere :S If I can get the formula for this one, I think I can solve the problem/
Where did you get the formula V= 2/3 pi r
3h? That can't be a volume, it has units of (length)
4! You are told that the tank consists of a cylinder of length h, radius r, and two hemis-spheres on the ends of radius r. The cylinder has volume \pi r^2 h and the two hemispheres, making one sphere, together have volume \frac{4}{3}\pi r^3. The volume of the tank is
V= \pi r^2 h + \frac{4}{3}\pi r^3= 200m.
The area of the curved surface of the cylinder is the circumference of the circle times length: 2\pi rh and the surface area of the two hemispheres, the area of one sphere, is \frac{4}{3}\pi r^2. Since the spherical parts cost twice as much per unit area, if we take the cost of the cylindrical part to be '1 per square foot', the cost of the whole thing will be
C= 2\pi rh+ \frac{8}{3}\pi r^2.
Solve the volume equation for h as a function of r, substitute into the cost equation and differentiate.
There is a slight complication: "The maximum length tank that can be safely transported to clients is 16 m long." If the solution you get above has the entire length of the tank, h+ 2r<= 16, you are done. If it has h+2r> 16, use h+2r= 16, so solve h+ 2r= 16 simultaneously with V= \pi r^2 h + \frac{4}{3}\pi r^3= 200m, ignoring the cost. That will not minimize cost but is "required by law"!
Orthodontist, you say
height is just the diameter of the hemisphere, or twice the radius. The formula for the volume is the following:
I assume you mean the height of the cylindrical part. If that's true then I am completely misinterpreting the problem! How did you arrive at that? Oh, wait! Do you mean the "height" of the tank when it is lying on it's side? If so, is that relevant to the problem?
A truck crossing the prairies at a constant speed of 110 km/h gets 8 km/ L of gas. Gas costs $0.68/L. The truck loses 0.10 km/L in fuel efficiency for each km/h increase in speed. Drivers are paid $35/h in wages and benefits. Fixed costs for running the truck are $15.50/h. If a trip of 450 km is planned, what speed will minimize operating expenses?
For this one, I tried to form a relationship between the speed, time and fuel but end up confusing myself. Am I supposed to take it 1 step at a time or what? I never studied this kind of optimization problem before.
A truck crossing the prairies at a constant speed of 110 km/h gets 8 km/ L of gas.The truck loses 0.10 km/L in fuel efficiency for each km/h increase in speed.
The second sentence tells you that fuel efficientcy (FE) in km/L is a linear function of v in km/h with slope -0.10 for v> 110. That is, it must be of the form FE= -.10(v-110)+ b. From the first sentence, FE= 8= -.10(110-110)+ b so b= 8. FE= -.10(v-110)+ 8= -.10v+ 11+8= -.10v+19. The volume of gas used for a trip of 450 miles, then, is (450)(-.10v+ 19)= -45v+ 8550 L and, with fuel costing 0.68 per Liter (how OLD is this problem??) the cost of fuel for the trip is (-45v+ 8550L)(0.68)= -30.6v+ 5814. At speed v, the time required for the trip will be (distance divided by speed) 450/v hours. Since drivers are paid $35 per hour, and the cost of running the truck is $15.50 per hour, the total cost for the truck per hour is $50.50 per hour and the total cost for the trip, excluding fuel, will be 50.50(450/v)= 22725/v.
The total cost of the trip, at speed v> 110 is
-30.5v+ 5814+ 22725/v. That's the function you want to minimize.
