Math Is Fun Forum

naturewild

Member

Registered: 2005-12-04

Posts: 30

Help with an optimization.. very wierd!

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

So I derive the relationship between R and H to this:

300 = (pie)(r²)(h)

But I dont know how to find the minimum price.

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?

This one, I'm simply lost at the variables.

Last edited by naturewild (2006-07-16 18:40:50)

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Help with an optimization.. very wierd!

Make a fuel tank volume equation, note both ends make a sphere volume, then middle cylinder.
Also make a separate area equation with ends multiplied by 2, so it is a cost equation, to be minimized.
Next analyze the 2 equations and decide what to do next.

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igloo myrtilles fourmis

DASET

Member

Registered: 2006-07-13

Posts: 5

Re: Help with an optimization.. very wierd!

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

This is an engineering economics type problem. You are given the total volume of the tank:
(1) 200 = pi r r h + pi 4 r r r/3
where r is the unknown radius and h is the unknown height of the cylinder.

You are given the inequality:
(2) 16 > 2 r + h

You are given that Xc (the cost per unit area of the cylinder) is half of Xs (the cost per unit area of the hemispheres). Hence, the total cost is:
(3) (Ac) Xc + (As) Xs = (2 pi r h) Xs/2 + (4 pi r r) Xs = C

At this point, we want to replace the height h in equation (3). Rearranging equation (1), we get:
(1)' h = 200/(pi r r) - 4 r/3  Replacing h in equation (3) and simplifying, we get:
(4) C = Xs pi [8 r r/3 + 200/(pi r)]

Equation (4) is the cost equation in terms of the unknown radius r. This is a smooth curve with r as the independent variable, so we can apply the calculus. In order to find the minimum cost, we take the derivative of equation (4) with respect to r and set the resulting expression to zero. Thus, we get:
(5)  dC/dr = Xs pi [16 r/3 - 200/(pi r r)] = 0

Solving equation (5) for r yields the value:
r r r = 200 3/(pi 16) and the cube root yields the desired value r = 2.285 m or 228.5 cm.
Substituting this value for r into equation (1)' yields the value of h = 9.142 m or 914.2 cm.

You may verify that this solution satisfies the inequality (2) and equation (1). To convince yourself that this value of r yields the minimum cost of manufacturing the tank, substitute r = 2.285 into the expression within the square brackets of equation (4). Then repeat using other values of r in the expression, say r = 2 and r = 3.

cheers

Last edited by DASET (2006-07-18 19:13:21)

All_Is_Number

Member

Registered: 2006-07-10

Posts: 258

Re: Help with an optimization.. very wierd!

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?

This one, I'm simply lost at the variables.

Let M = the cost due to fixed hourly operating/maintenance expenses of the truck,

F = cost of fuel, L = labor cost, C = total cost, and t = time it takes to make the trip,

such that:  M = $15.50 * t, F = $.68 * 450 / (8 - .1 * (450 / t - 110)), L = $35.00 * t, and C = M + F + L.

Notice that 8 - .1 * (450 / t - 110) yields a negative solution for all values of t < 45 / 19. Since the vehicle cannot get negative fuel efficiency, we are not interested in any solutions for t = 45 / 19.

C = 15.50 * t + .68 * 450 / (8 - .1 * (450 / t - 110)) + 35.00 * t

dC/dt = 15.5 - 13 770 / (19 * t - 45)^2 + 35

Solving for dC/dt = 0 yields t = 3.23752

450 / 3.23752 = 138.995

Operating expenses are minimized if the driver drives at about 139 kilometers per hour.

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You can shear a sheep many times but skin him only once.