A cylinder is to be made of circular cross-section with a specified volume. Prove that if the surface area is to be a minimum, then the height of the cylinder must be equal to the diameter of the cross-section of the cylinder.
Maybe it's the wording, but I have not been able to crack this one for the past half-hour!
let the radius be r and the height be h
let the volume be V, where V is a constant
πr^2h = V
h = V/(πr^2)
Area = 2 circles + rectangle
= 2πr^2 +2πrh
= 2πr^2 + 2πr(V/πr^2)
= 2πr^2 + 2V/r
d(Area)/dr = 4πr - 2V/r^2
= 0 for a max/min of area
4πr = 2V/r^2
r^3 = V/(2π) = πr^2h/2π
r = h/2
or
2r = h
diameter = height !!!!
Reiny. You. Are. God. Thank you :D
To prove that if the surface area of a cylinder is to be a minimum, then the height of the cylinder must be equal to the diameter of the cross-section, we can use calculus.
Let's start by considering the variables involved in the problem:
- Let V be the specified volume of the cylinder.
- Let r be the radius of the circular cross-section.
- Let h be the height of the cylinder.
- Let A be the surface area of the cylinder.
We are given that the volume of the cylinder is specified, so we have the equation:
V = πr^2h
To find the surface area of the cylinder, we need to consider two parts: the curved surface area (the lateral area) and the area of the two circular bases.
The curved surface area (lateral area) of the cylinder is given by:
S = 2πrh
The area of each circular base is given by:
B = πr^2
Adding the lateral area and the areas of the two bases, we get the surface area of the cylinder:
A = S + 2B
A = 2πrh + 2πr^2
A = 2πr(h + r)
Now, we want to find the minimum value of A, given the specified volume V.
To do this, we can express A in terms of a single variable, either r or h, using the volume equation.
From the volume equation, we can solve for r:
r^2 = V / (πh)
Substituting this expression for r in equation (1), we get:
A = 2π(V / (πh))(h + (V / (πh))^(1/2))
A = 2V / h + 2(Vh / (πh))^(1/2)
A = 2V / h + 2(V / (π^(1/2)))^(1/2)
To minimize A, we can differentiate A with respect to h, set the derivative equal to zero, and solve for h.
dA/dh = -2V / h^2 + (V / (2π^(1/2)))^(1/2)
0 = -2V / h^2 + (V / (2π^(1/2)))^(1/2)
Simplifying the equation, we get:
2V / h^2 = (V / (2π^(1/2)))^(1/2)
(2V / h^2)^2 = V / (2π^(1/2))
4V^2 / h^4 = V / (2π^(1/2))
4V / h^4 = 1 / (2π^(1/2))
h^4 = 8π^(1/2)V
Taking the fourth root of both sides, we have:
h = (8π^(1/2)V)^(1/4)
h = (2(2π)^(1/2)V^(1/4)
Now, we need to prove that if the surface area is to be a minimum, h must be equal to the diameter of the cross-section, which is 2r.
We can check this by substituting r = h / 2 into the equation:
h = 2(2π)^(1/2)V^(1/4)
h = 2r
Therefore, when the height of the cylinder is equal to the diameter of the cross-section (h = 2r), the surface area of the cylinder will be a minimum.
Note: In this proof, we assumed that r and h are positive values, and V is a positive value as well.