PaulKraemer
New member
- Joined
- Apr 10, 2011
- Messages
- 45
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
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