Calculus Question

[silverpig]

Oh, and:

"Yes, it is possible that there is a point of inflection where there is an extremum" is BS too if you take extremum = local max/min and not just extremum = [f' = 0] 🙂

 

[PhaZe]

Originally posted by: MrCodeDude
A piece of wire 10m long is cut into two pieces. One piece is bent into a squre and the other is bent into an equilateral triangle. How should the wire be cut so that the total area enclosed is: (a) a maximum (b) a minimum

Wouldn't the square produce the most area and the triangle would produce the least area, but there is some mathematical way to do this.

quick reminder of what the original topic was.
😛

 

[Whitecloak]

Originally posted by: MrCodeDude

Originally posted by: agnitrate
f (x) = e^x - e^(-x)
f ' (x) = e^x + e^(-x)
f '' (x) = e^x - e^(-x)

Wouldn't the first deriv be x*(e^(x-1))+x*(e^(-x-1))?

MCD, dont you love me? 🙁

 

[RaynorWolfcastle]

Originally posted by: silverpig

Originally posted by: RaynorWolfcastle
OK, since everyone is going nuts over semantics I'll clear up what I meant:

if f'(x) = 0 you have a local maximum or minimum because it means that the tangent is HORIZONTAL.
GENERALLY If f'(x) = 0 it implies absolutely nothing about that point being an inflection point; it is the second derivative test that gives you information about that (f"(x) = 0).

If you want, I can draw you all a nice little function that has n points of inflection without having either a max or a min (using high order polynomial).

Yes, it is possible that there is a point of inflection where there is an extremum, but by the logic you're using I could say "y = x^2, has a zero at x = 0 therefore all functions have a zero when the dependent variable is zero". That is why I said he was wrong, the statement was wrong because there is no GENERAL relation between the first derivative test and the points of inflection.

I'm done with this thread, I don't need to be taught elementary calculus, I understand it very well thank you very much.

Actually, you fvcked this explanation up too. 🙂

"if f'(x) = 0 you have a local maximum or minimum because it means that the tangent is HORIZONTAL."

BS. The x^3 example. You have f' = 0, but no local max or min. Tangent is horizontal.

Unless of course you meant to limit the scope of your statement to the problem stated by the original poster. I don't usually nitpick like this, but I really don't want MrCodeDude to get confused with half correct information or unclear statements.

You're right, I meant you have to apply the first derivative test which means finding the points of horizontal tangency and verifiying whether or not the sign of f'(x) changes.

You're also right about your second statement too I was just following up on my original, false, statement.

Thanks for correcting my mistakes, when/if I ever do this stuff now I do it intuitively without any formality but I should have been more careful in explaining it to someone else. My apologies Mr. CodeDude, but my point really was regarding the fact that the second derivative test is not really related to the first derivative test.

 

[MegaloManiaK]

No way dude, the death star would so own the enterprise in an all out battle.

😛

 

[sputkin]

Total Area = (x/4)*(x/4) + sqrt(3)/4 * (10-x)/3 * (10-x)/3

This equation looks good to me. Did anyone do the actual number crunching?
I get the min area as 2.72 where x=4.35 . Since it's a parabola, there is no relative or global max, but remember the interval for x is [0,10] so for our particular problem, the max area is 6.25 where x=10 (since the endpoints counts as an extrema).

Sooo.... the min area occurs when 4.35m of the wire is used for the square and 5.65m of it is used for the triangle. The max when all the wire is used for the square.

Ain't Calc grand??