A company wants to build a cylindrical container with a semi-sphere lid.
For a fixed volume V , the company wants to use a minimal amount of material for container
and lid combined. Which radius r and height h of the container minimize the surface area for
container and lid combined?
v = πr^2h + 2/3 πr^3
so, h = (v - 2/3 πr^3)/πr^2
= v/(πr^2) - 2r/3
the surface area is
a = πr^2 + 2πrh + 2πr^2
= 3πr^2 + 2πr(v/(πr^2) - 2r/3)
= 5πr^2/3 + 2v/r
da/dt = 10πr/3 - 2v/r^2
= (10πr^3/3 - 2v)/r^2
da/dt=0 when
10πr^3/3 - 2v = 0
r^3 = 3v/(5π)
and express h in terms of that!
Great, thank you Steve!
To minimize the surface area of the container and lid combined for a fixed volume V, we can use the process of calculus.
Let's start by determining the formulas for the surface areas of the cylinder and the semi-sphere:
1. Surface Area of the Cylinder:
The formula for the surface area of a cylinder is given by A_cylinder = 2πrh + πr^2, where r is the radius and h is the height.
2. Surface Area of the Semi-Sphere:
The formula for the surface area of a semi-sphere is given by A_semi-sphere = 2πr^2, where r is the radius.
Now, since the company wants to minimize the combined surface area of the container and lid, we can express it as a function of one variable. Let's define S as the combined surface area:
S(r, h) = 2πrh + πr^2 + 2πr^2
Here, the first term represents the surface area of the cylindrical part, the second term is the area of the circular bottom of the cylinder, and the third term is the surface area of the semi-sphere lid.
Since we want to minimize S, we need to find the values of r and h that minimize the function.
To do this, let's differentiate S with respect to r and h, and then set the derivatives equal to zero.
∂S/∂r = 2πh + 2πr + 4πr = 2π(h + 3r) = 0
∂S/∂h = 2πr = 0
Setting these derivatives equal to zero, we find two equations:
1. Equation 1: h + 3r = 0
2. Equation 2: r = 0
From Equation 2, we can see that if r = 0, the surface area will be zero. However, a container and lid with zero radius will not hold any volume, so it is not a relevant solution.
Now, let's solve Equation 1 for r:
h + 3r = 0
3r = -h
r = -h/3
We should note that h and r must be positive values. Since we are assuming V is a fixed positive volume, h must be a positive value, and thus r will also be positive.
Now that we have the value of r in terms of h, we can substitute it back into the equation for the surface area S:
S(h) = 2πh(-h/3) + π(-h/3)^2 + 2π(-h/3)^2
Simplifying this equation will give us the minimized surface area function S(h).
Finally, to find the minimum surface area, we can differentiate S(h) with respect to h and set it equal to zero, and then solve for h. Once we have the value of h, we can substitute it back into the equation for r to find the corresponding radius.