Tricky Derivative Question (1 Viewer)

[inedible]

Consider the curve y = e^(-x^2) and we drew a rectangle bound by the curve and the x axis.

Show that the rectangle will have maximum area when the upper right hand side corner is a point of inflexion.

And can you include working out please?

Sorry I don't have a picture and thanks for your help!

 

[sydneyroosters]

Area of rectangle = (2x) x e^(-x^2)
= 2xe^(-x^2)

dA/dx = 2e^(-x^2) + 2x(-2x)e^(-x^2) [chain rule]
= 2e^(-x^2) - (4x^2)e^(-x^2)

max/min at dA/dx = 0

2e^(-x^2) - (4x^2)e^(-x^2) = 0
e^(-x^2) [2 - 4x^2] = 0
e^(-x^2) cannot = 0 since -x^2 cannot = ln 0

therefore 4x^2 = 2
x^2 = 1/2
x = 1/root 2 since we're considering distance

d2A/dx2 = -4xe^(-x^2) - 8xe^(-x^2) + (8x^3)e^(-x^2)
= e^(-x^2) [-4x - 8x + 8x^3]
= e^(-x^2) [-12x + 8x^3]

at x = 1/root 2, d2A/dx2 = -3.43, therefore max at 1/root2

now consider y = e^(-x^2)

dy/dx = -2xe^(-x^2)
d2y/dx2 = -2e^(-x^2) + (4x^2)e^(-x^2)
inflexion at d2y/dx2 = 0

-2e^(-x^2) + (4x^2)e^(-x^2) = 0
e^(-x^2)[-2 + 4x^2] = 0
4x^2 = 2
x = 1/root2

at x = 0.6, d2y/dx2 = -ve
at x = 1/root2, d2y/dx2 = 0
at x = 0.8, d2y/dx2 = +ve

therefore inflexion at 1/root2

Thus, maximum area of rectangle occurs when upper corner is at inflexion

p.s. sorry if there is some tautology in the working

 

[emmcyclopedia]

-2e^(-x^2) + (4x^2)e^(-x^2) = 0
e^(-x^2)[-2 + 4x^2] = 0
4x^2 = 2
x = 1/root2

at x = 0.6, d2y/dx2 = -ve
at x = 1/root2, d2y/dx2 = 0
at x - 0.8 = +ve

therefore inflexion at 1/root2

^^ nice job =]
would just like to add that I don't think you'd need to show concavity of the point of inflexion, y''=0 is enough.

 

[Trebla]

You should check change in concavity in the neighbourhood to confirm a point of inflexion. The second derivative being zero is not adequate.

For example, y = x⁴
dy/dx = 4x³
d²y/dx² = 12x²
This seems to imply that there is a point of inflexion at x = 0. However, note that d²y/dx² > 0 everywhere else which means it is always concave up. There is no change in concavity so hence x = 0 is NOT a point of inflexion even though it satisfies d²y/dx² = 0

 

[cutemouse]

You should check change in concavity in the neighbourhood to confirm a point of inflexion. The second derivative being zero is not adequate.

For example, y = x⁴
dy/dx = 4x³
d²y/dx² = 12x²
This seems to imply that there is a point of inflexion at x = 0. However, note that d²y/dx² > 0 everywhere else which means it is always concave up. There is no change in concavity so hence x = 0 is NOT a point of inflexion even though it satisfies d²y/dx² = 0

Or use the 3rd derivative test (easier).

 

[Trebla]

In the above example, the third derivative at x = 0 is also zero. What can be concluded from that?

 

[cutemouse]

In the above example, the third derivative at x = 0 is also zero. What can be concluded from that?

That x=0 is NOT a point of inflexion. For a point of inflexion the third derivative cannot equal zero.

Obviously you'd only use this method when the third derivative is easy to obtain (like in your example).

 

[buchanan]

For a point of inflexion the third derivative cannot equal zero.

This is not correct. Consider y=x⁵. Third derivative is 0 at (0,0), but this IS an inflection.

 

[cutemouse]

This is not correct. Consider y=x⁵. Third derivative is 0 at (0,0), but this IS an inflection.

lol, that method I've used always works for cubics... Not too sure if the leading power is greater than that.

What would your thoughts be on this method?

 

[buchanan]

Well for a cubic, the third derivative will always be a non-zero constant, hence for cubics, the third derivative test will always work.

Don't let this make you think it will always work for all functions though. The third derivative test is sufficient but not necessary.

 

[cutemouse]

Well for a cubic, the third derivative will always be a non-zero constant, hence for cubics, the third derivative test will always work.

Don't let this make you think it will always work for all functions though. The third derivative test is sufficient but not necessary.

Umm... okay... so it doesn't work with all functions?

 

[buchanan]

I made a thread on it here

 

[emmcyclopedia]

oops, mybad. i retract my original post!

 

[random-1005]

Or use the 3rd derivative test (easier).

make up your own tests , good job noob

 

[inedible]

Thanks! You guys are so helpful.