Calculus Question

[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.

 

[dighn]

i think by total area enclosed it means area of square + area of triangle

 

[Heisenberg]

Express the area as the sum of the triangle and square in terms of the side lengths and then take the derivative. Set it equal to zero, and find the correct side lengths.

 

[jacob0401]

 

[Howard]

Originally posted by: Heisenberg
Express the area as the sum of the triangle and square in terms of the side lengths and then take the derivative. Set it equal to zero, and find the correct side lengths.

You'd have to explain to him why that works.

 

[duragezic]

Points at which deriv = zero is a minimum or a maximum.

 

[xirtam]

Vote for homework forum!

 

[MikeMike]

its all of one or the other i believe.

MIKE

 

[Shooters]

Originally posted by: eagle
Points at which deriv = zero is a minimum or a maximum.

or an inflection point

 

[jacob0401]

Originally posted by: Shooters

Originally posted by: eagle
Points at which deriv = zero is a minimum or a maximum.

or an inflection point

nah...inflection point is where the graph changes from concave up or concave down and is the zeroes of the second derivative

 

[Whitecloak]

let x be the perimeter of the square -> 10-x is the perimeter of the triangle.
Therefore, length of a side of the square = x/4 and length of a side of the equilateral triangle = (10-x)/3

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

differentiatiate the above equation and solve for zero to get the inflexion point. Differentiate once again to find out whether the point is a maxima or a minima.

 

[Shooters]

Originally posted by: jacob0401

Originally posted by: Shooters

Originally posted by: eagle
Points at which deriv = zero is a minimum or a maximum.

or an inflection point

nah...inflection point is where the graph changes from concave up or concave down and is the zeroes of the second derivative

First derivative = 0 means the tangent line has zero slope (i.e. horizontal); this occurs at an inflection point as well as at mins and maxes.

 

[RaynorWolfcastle]

Originally posted by: Shooters

Originally posted by: jacob0401

Originally posted by: Shooters

Originally posted by: eagle
Points at which deriv = zero is a minimum or a maximum.

or an inflection point

nah...inflection point is where the graph changes from concave up or concave down and is the zeroes of the second derivative

First derivative = 0 means the tangent line has zero slope (i.e. horizontal); this occurs at an inflection point as well as at mins and maxes.

no, reread your math book, jacob is right

 

[Shooters]

Originally posted by: RaynorWolfcastle

Originally posted by: Shooters

Originally posted by: jacob0401

Originally posted by: Shooters

Originally posted by: eagle
Points at which deriv = zero is a minimum or a maximum.

or an inflection point

nah...inflection point is where the graph changes from concave up or concave down and is the zeroes of the second derivative

First derivative = 0 means the tangent line has zero slope (i.e. horizontal); this occurs at an inflection point as well as at mins and maxes.

no, reread your math book, jacob is right

I'm not trying to be an arrogant prick, but I know what I'm talking about. Jacob is right about the fact that an inflection point indicates a change in concavity, but the first derivative does equal zero at an inflection point just as it does at a min or max.

For example, if f(x) = x^3, then f'(x) = 3x^2. Therefore, f'(0) = 3(0)^2 = 0. Look at the graph of f(x) = x^3 and you will see that there is neither a min or max at x = 0; it is an inflection point.

inflection point
stationary point
first derivative test

Next time make sure you're right before you correct somebody.

 

[agnitrate]

Originally posted by: Shooters

I'm not trying to be an arrogant prick, but I know what I'm talking about.

f (x) = e^x - e^(-x)
f ' (x) = e^x + e^(-x)
f '' (x) = e^x - e^(-x)

f ' (0) = 2
f '' (0) = 0

No you don't, unless I did something horribly wrong in my sick state.

-silver

 

[Oscar1613]

n/m too slow

 

[silverpig]

Originally posted by: Shooters

Originally posted by: RaynorWolfcastle

Originally posted by: Shooters

Originally posted by: jacob0401

Originally posted by: Shooters

Originally posted by: eagle
Points at which deriv = zero is a minimum or a maximum.

or an inflection point

nah...inflection point is where the graph changes from concave up or concave down and is the zeroes of the second derivative

First derivative = 0 means the tangent line has zero slope (i.e. horizontal); this occurs at an inflection point as well as at mins and maxes.

no, reread your math book, jacob is right

I'm not trying to be an arrogant prick, but I know what I'm talking about. Jacob is right about the fact that an inflection point indicates a change in concavity, but the first derivative does equal zero at an inflection point just as it does at a min or max.

For example, if f(x) = x^3, then f'(x) = 3x^2. Therefore, f'(0) = 3(0)^2 = 0. Look at the graph of f(x) = x^3 and you will see that there is neither a min or max at x = 0; it is an inflection point.

inflection point
stationary point
first derivative test

Next time make sure you're right before you correct somebody.

You guys are BOTH WRONG.

It can be either way. You are right about the x^3 part. The derivative is zero and there is an inflection point, but this is not always the case. Try sin(x) for example. There is an inflection point at x=0, the second derivative is 0, but the first derivative is 1.

 

[EmperorIQ]

Originally posted by: silverpig

Originally posted by: Shooters

Originally posted by: RaynorWolfcastle

Originally posted by: Shooters

Originally posted by: jacob0401

Originally posted by: Shooters

Originally posted by: eagle
Points at which deriv = zero is a minimum or a maximum.

or an inflection point

nah...inflection point is where the graph changes from concave up or concave down and is the zeroes of the second derivative

First derivative = 0 means the tangent line has zero slope (i.e. horizontal); this occurs at an inflection point as well as at mins and maxes.

no, reread your math book, jacob is right

I'm not trying to be an arrogant prick, but I know what I'm talking about. Jacob is right about the fact that an inflection point indicates a change in concavity, but the first derivative does equal zero at an inflection point just as it does at a min or max.

For example, if f(x) = x^3, then f'(x) = 3x^2. Therefore, f'(0) = 3(0)^2 = 0. Look at the graph of f(x) = x^3 and you will see that there is neither a min or max at x = 0; it is an inflection point.

inflection point
stationary point
first derivative test

Next time make sure you're right before you correct somebody.

You guys are BOTH WRONG.

It can be either way. You are right about the x^3 part. The derivative is zero and there is an inflection point, but this is not always the case. Try sin(x) for example. There is an inflection point at x=0, the second derivative is 0, but the first derivative is 1.

and we have a winner!!!

also, in one of the webpaged linked it states
"The first derivative test can sometimes distinguish inflection points from extrema for differentiable functions "
note the sometimes?

 

[DXM]

Originally posted by: silverpig

Originally posted by: Shooters

Originally posted by: RaynorWolfcastle

Originally posted by: Shooters

Originally posted by: jacob0401

Originally posted by: Shooters

Originally posted by: eagle
Points at which deriv = zero is a minimum or a maximum.

or an inflection point

nah...inflection point is where the graph changes from concave up or concave down and is the zeroes of the second derivative

First derivative = 0 means the tangent line has zero slope (i.e. horizontal); this occurs at an inflection point as well as at mins and maxes.

no, reread your math book, jacob is right

I'm not trying to be an arrogant prick, but I know what I'm talking about. Jacob is right about the fact that an inflection point indicates a change in concavity, but the first derivative does equal zero at an inflection point just as it does at a min or max.

For example, if f(x) = x^3, then f'(x) = 3x^2. Therefore, f'(0) = 3(0)^2 = 0. Look at the graph of f(x) = x^3 and you will see that there is neither a min or max at x = 0; it is an inflection point.

inflection point
stationary point
first derivative test

Next time make sure you're right before you correct somebody.

You guys are BOTH WRONG.

It can be either way. You are right about the x^3 part. The derivative is zero and there is an inflection point, but this is not always the case. Try sin(x) for example. There is an inflection point at x=0, the second derivative is 0, but the first derivative is 1.

Owned. hahahahaha

 

[Shooters]

Yeah, you guys are right about the fact that f'(x) = 0 isn't a necessary condition for an inflection, but it is possible. The basic jist of the story is that f'(x) = 0 doesn't necessarily mean that you have a min or max.

 

[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.

 

[duragezic]

No, you use the fact that points at which f`(x) = 0 is a min or max, and these values are convientenly the answers to the minimum and maximum dimensions of such and such shape so that the volume/perimeter/area, etc... is maximized/minimized. It is like proving that the answers you get for length/width/whatever would indeed produce a max or min area.

 

[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))?

 

[Oscar1613]

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))?

no... e^x is a special case you'll learn about. f'(e^x)=e^x and f'(e^ax)=ae^ax

 

[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.