# Optimization height and radius problems

1. Jul 17, 2006

### coldcell

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/

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.

2. Jul 17, 2006

### 0rthodontist

Well in the first part the height is just the diameter of the hemisphere, or twice the radius. The formula for the volume is the following:
(voume) = (volume of hemispherical caps) + (volume of cylindrical body)
Think in terms of the variables r, the radius of the caps and the cylindrical body, and L - 2r, which is the length of the cylindrical body disregarding the caps if L is the total length.

For pricing it, you need to arrange it so that the cost of the caps per unit area is twice the cost of the cylindrical body per unit area. Any pricing scheme that satisfies that is enough for this problem.

In the second one, you know that
(total cost of trip) = (time of trip) * (fixed hourly costs and wages) + (cost of fuel)
Now you need to find (cost of fuel) in terms of the speed, v.

3. Jul 17, 2006

### HallsofIvy

Staff Emeritus
Where did you get the formula V= 2/3 pi r3h? 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
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?

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.

Last edited: Jul 17, 2006