(Thank you for any help because I am not good at setting up word problems)
What are the dimensions that will minimize the amount of material needed to manufacture a standard oil drum that is in the shape of a cylinder, with closed top and bottom. The drum must have a volume of 2000pi cubic inches. (Again thank you for any help)
let the radius be r
and let the height be h
πr^2 h = 2000π
h = 2000/r^2
we want to minimize the surface area A
A = 2πr^2 + 2πrh
= 2πr^2 + 2πr(2000/r^2)
= 2πr^2 + 4000π/r
dA/dr = 4πr - 4000π/r^2
= 0 for a minimum of A
4πr = 4000π/r^2
r = 1000/r^2
r^3 = 1000
r = 10
then h = 2000/100 = 20
the height should be 20 inches, and the radius should be 10 inches.
Notice the height = diameter for a minimum surface area of a cylinder.
To find the dimensions that will minimize the amount of material needed to manufacture the oil drum, we need to consider the volume and surface area of the cylinder.
Let's start by defining the variables:
- r = radius of the cylinder base
- h = height of the cylinder
- V = volume of the cylinder
The volume of a cylinder is given by the formula V = πr²h. In this problem, we are given that the volume of the cylinder should be 2000π cubic inches. Therefore, we have the equation:
2000π = πr²h
To minimize the amount of material needed, we need to minimize the surface area of the cylinder. The surface area of a cylinder can be calculated using the formula A = 2πrh + 2πr².
Since we want to minimize the surface area, we can substitute the value of h from the volume equation into the surface area equation:
A = 2πrh + 2πr² [Substituting h from the volume equation]
Now we can express the surface area (A) solely in terms of r:
A = 2πr(2000/πr²) + 2πr²
A = 4000/r + 2πr²
To find the minimum surface area, we need to find the minimum value of A. To do this, we can take the derivative of A with respect to r, set it equal to zero, and solve for r:
dA/dr = -4000/r² + 4πr
Since we want to find the minimum value, we need to find the critical points by setting the derivative equal to zero:
0 = -4000/r² + 4πr
4000/r² = 4πr
1000/r² = πr
1000 = πr³
r³ = 1000/π
r = (1000/π)^(1/3)
Now that we have the value of r, we can substitute it back into the volume equation to find the corresponding height (h):
2000π = π((1000/π)^(1/3))²h
2000 = (1000/π)^(2/3)h
h = 2000/((1000/π)^(2/3))
Therefore, the dimensions that will minimize the amount of material needed to manufacture the oil drum are:
- r = (1000/π)^(1/3)
- h = 2000/((1000/π)^(2/3))
Be sure to simplify and calculate the values for r and h using a calculator.