Maximum and Minimum

1. Jul 6, 2013

Jason298

Can someone help me as a am racking my brain. I have been given this question;

Two oil pipes with circular cross-sections are subject to the constraint that the sum of their
diameters is 1m. By using careful reasoning find the diameters that give the maximum and
minimum possible combined cross-sectional areas. In each case give the combined crosssectional
area.

Which I am unsure of. I dont get how you can get a maximum and minimum of a circle diameter when the volume of material going through it is irrelevant.

Any assistance would be greatly appreciated.

2. Jul 6, 2013

phinds

Can you not find a way to start attacking the problem mathematically? What is the area of a circle in terms of what you know?

3. Jul 7, 2013

Jason298

I believe I have cracked it, however will not know until it has been submitted so still need to try different angles to see what else I can come up with.

4. Jul 8, 2013

verty

The question has nothing to do with angles. This is a basic optimization problem, your textbook will explain how to solve them.

5. Jul 8, 2013

phinds

I take it you are not a native English speaker. "Angles" in this context has nothing to do with geometry, it mean "different ways to try something".

6. Jul 13, 2013

corsix049

So how did you work it out then? I can't seem to figure out how to solve your problem mathematically.

7. Jul 13, 2013

Millennial

You want to maximize $\pi(r_a^2 + r_b^2)$ based on the constraint that $r_a + r_b = 1$.
Try starting with the identity $(r_a+r_b)^2 = r_a^2 + r_b^2 + 2r_ar_b$.

8. Jul 13, 2013

verty

This is a typical question that any calculus book will show you how to solve, usually under "maxima and minima" or "applications of the derivative". I would class it as an easy question if you know the method, which is why I said above that the poster should refer to his textbook for the method of solving these.

We were all waiting for the poster to present some work because, if he knew the method, he would have been able to get started. If he got stuck somewhere, we could have helped. But being unable to start means he does not know enough to benefit from our help. And, it's against the rules.

If the OP can show some work, I'm sure he will receive much help.

9. Jul 13, 2013

haruspex

Let one of them have diameter x. What's the diameter of the other? What's the sum of the cross-sectional areas?

10. Jul 13, 2013

verty

I think the question following this will be, what do I do next? And the only answer one can give to that is to explain in general how to solve these. I was hoping we could avoid that.

Actually, I learned enough in school to solve this with just the most basic calculus knowledge. This isn't a big deal really, I shouldn't worry.

Last edited: Jul 13, 2013
11. Jul 13, 2013

Millennial

It is solvable the way I gave without any general knowledge. Just substitute $r_a = 1-r_b$ or vice versa after arrangement.

12. Jul 13, 2013

Jason298

I started with the Area, substituted R for x and y. Worked out the starting area which was my maximum. Transposed the diameter formula to give me Y=1-X
Differentiated the transposed area formula using the (1-X) which got rid of Y, so eventually I got x=0.5, I put this in the original area formula which gave me my minimum. I then done my graph with x=1 Y=0 (diameter) and went through x=.9 y=.1, x=.8 y=.2 and so on. This gave me my changing areas. This proved my answer and gave me a 'U' like graph.

13. Jul 13, 2013

SteamKing

Staff Emeritus
This is incorrect. According to the OP, the DIAMETERs must sum to 1 meter. Then, the sum of the areas of the two pipes should be expressed in terms of their diameters

14. Jul 13, 2013

Ray Vickson

The problem as written has no solution, as you can have area =+infinity by taking radii of + and - infinity. Restricting radii to non-negative values gives a solvable problem, but not one to which calculus methods apply.

15. Jul 13, 2013

verty

Pipes cannot have negative radii. At least no pipe I've ever seen. Perhaps a Klein bottle can, I don't know about that.

16. Jul 13, 2013

Ray Vickson

Of course, and that is why your formulation was incomplete--it needed the sign restriction. Just because you and I know the variables must be >= 0 does not mean that a solver package would know it. If you were to submit your problem *exactly as written* to, eg., the Excel Solver it would essentially give you an unbounded solution. Sometimes we do not need to specify sign restrictions (eg., for the present case with *minimization*) but in other times we need to be more explicit.

Anyway, in this example the maximum and minimum problems are completely different, even in the presence of sign-restricted radii. Calculus methods solve the min problem, but not the max problem.

Last edited: Jul 13, 2013
17. Jul 13, 2013

HallsofIvy

Staff Emeritus
Or use $r_1+ r_2= 1/2$.

18. Jul 13, 2013