derivative word problem

[PaulKraemer]

Hi,

I am having trouble with the following problem:

Find the altitude of the right circular cylinder of maximum curved surface area that can be inscribed in sphere of radius a.

..what I have done so far is.....

(1) Let h be the height of the cylinder and r be the radius of the cylinder.
(2) Relating r to h, I got r = sqrt [ a^2 - (1/2)h^2 ]
(3) Find formula for curved surface area A:

A = 2* pi * r * h = 2 * pi * h * sqrt [ a^2 - (1/2)h^2 ]

(4) Find derivative of A:

A' = [(2 * pi) * (a^2 - h^2)] / sqrt [(a^2 - (1/2)h^2]

(5) Find non-negative critical numbers for A':

h = a
h = sqrt(2)a

(6) Find second derivative A'':

A'' = [ (pi * h^3) - (3 * pi * h * a^2) ] / sqrt [(a^2 - (1/2)h^2]

(7) Use second derivative test to determine if there is a maximum at either h = a or at h = sqrt(2)a

A''(a) = (-2 * pi * a^3) / sqrt [(a^2 - (1/2)h^2] < 0
A"(sqrt(2)a) = (-sqrt(2) * pi * a^3) / sqrt [(a^2 - (1/2)h^2] < 0

The fact that the second derivate is negative at both h = a and h = sqrt(2)a leads me to believe that there is a local max at both h = a and h = sqrt(2)a.

(8) I figured I should plug both h = a and h = sqrt(2)a into the formula for curved surface area in step (3). When doing this, I realized that when h = sqrt(2)a, then r = sqrt [ a^2 - (1/2)h^2 ] = 0 and therefore the surface area A = 0.

For this reason, I figured the max must be at h = a. However, the back of the book says the answer is h = sqrt(2)a

I was just wondering if anyone here can see where I went wrong.

Thanks in advance,
Paul

 

[Deleted member 4993]

Hi,

I am having trouble with the following problem:

Find the altitude of the right circular cylinder of maximum curved surface area that can be inscribed in sphere of radius a.

..what I have done so far is.....

(1) Let h be the height of the cylinder and r be the radius of the cylinder.
(2) Relating r to h, I got r = sqrt [ a^2 - (1/2)h^2 ]
(3) Find formula for curved surface area A:

A = 2* pi * r * h = 2 * pi * h * sqrt [ a^2 - (1/2)h^2 ]

(4) Find derivative of A:

A' = 2* π * (2a² - h²)/√(4a² - h²)

A' = 0 → h = ±a * √2

A' = [(2 * pi) * (a^2 - h^2)] / sqrt [(a^2 - (1/2)h^2]

(5) Find non-negative critical numbers for A':

h = a
h = sqrt(2)a ...... how are you getting this from your equation?

(6) Find second derivative A'':

A'' = [ (pi * h^3) - (3 * pi * h * a^2) ] / sqrt [(a^2 - (1/2)h^2]

(7) Use second derivative test to determine if there is a maximum at either h = a or at h = sqrt(2)a

A''(a) = (-2 * pi * a^3) / sqrt [(a^2 - (1/2)h^2] < 0
A"(sqrt(2)a) = (-sqrt(2) * pi * a^3) / sqrt [(a^2 - (1/2)h^2] < 0

The fact that the second derivate is negative at both h = a and h = sqrt(2)a leads me to believe that there is a local max at both h = a and h = sqrt(2)a.

(8) I figured I should plug both h = a and h = sqrt(2)a into the formula for curved surface area in step (3). When doing this, I realized that when h = sqrt(2)a, then r = sqrt [ a^2 - (1/2)h^2 ] = 0 and therefore the surface area A = 0.

For this reason, I figured the max must be at h = a. However, the back of the book says the answer is h = sqrt(2)a

I was just wondering if anyone here can see where I went wrong.

Thanks in advance,
Paul

.

 

[Deleted member 4993]

Cody,

Paul was trying to maximize the curved surface only.

 

[galactus]

Oh, thanks SK. I reckon I had better read closer. No lid or bottom on the can.

Well, I will leave it there.

 

[PaulKraemer]

Thank you Subhotosh,

I made my mistake right in the beginning....my formula for r should have been:

r = sqrt [ a^2 - (1/4)h^2 ] instead of r = sqrt [ a^2 - (1/2)h^2 ]

With that change, I now get the same formula as you for A':

A' = 2* π * (2a² - h²)/√(4a² - h²)

As you noted, h = ±a * √2 is a critical number, but I rule out the negative. I calculated A'' as follows:

A'' = [ (-12 * π * h * a^2) + (2 * π * h^3) ] / [ (4a^2 - h^2)^(3/2)]

plugging (a * √2) into A'', I come up with:

A''(a * √2) = [ (-8) * √2 * π * a^3 ] / 2a^2 < 0

Thus, the second derivative test says that this is a local max....

My only question is...isn't h = 2a also a critical number of A' ? h = 2a makes the denominator of A'' = zero so I can't use the second derivative test on this one. If I am thinking straight, I can tell that h = 2a is not a max because r = sqrt [ a^2 - (1/4)h^2 ] = 0 when h = 2a and therefore...A = 2 * π * r * h = 0, so this can't be a max.

Is this the proper way to rule out h = 2a as a solution? Or is there a better way?

Thanks again,
Paul