a cylindrical soup can has a volume of 335cm^3.find the dimensions(radius r and height h)that minimise the surface area of such a can.
V = πr^2 h
h = 335/(π r^2)
SA = 2π r^2 + 2πrh
= 2πr^2 + 2πr(335/(πr^2)
= 2πr^2 + 670/r
d(SA)/dr = 4πr - 670/r^2
= 0 for a min of SA
4πr = 670/r^2
4πr^3 = 670
r = (670/(4π))^(1/3) = 3.763757606..
subbing back into h
h = 7.5275152..
notice that h : r = 2 : 1
that is, the height should be equal to the diameter for a minimum surface area
To minimize the surface area of the cylindrical soup can, we need to find the dimensions (radius r and height h) that will minimize it.
Let's start by writing the formulas for the volume and surface area of a cylinder:
1. Volume of a cylinder: V = πr^2h
2. Surface area of a cylinder: A = 2πrh + 2πr^2
Given that the volume of the can is 335 cm^3, we have: V = 335 cm^3
Now, we need to minimize the surface area A, so we will differentiate it with respect to r and set it equal to zero to find the critical points:
dA/dr = 2πh + 4πr = 0
Simplifying the equation, we get: h = -2r
We also know the volume of the can, so we can substitute the value of h in terms of r:
335 = πr^2(-2r)
Simplifying further, we get: 335 = -2πr^3
To solve for r, divide both sides by -2π and take the cube root:
r^3 = -335 / (2π)
r = [(-335 / (2π))^(1/3)]
Using this value of r, we can find h using the equation h = -2r:
h = -2 × [(-335 / (2π))^(1/3)]
You can use a calculator to compute the values of r and h approximately. Keep in mind that the height (h) must be positive since negative height does not make sense in this context.
Note: While the above steps provide the mathematical process, keep in mind that in practice, you may round the values of r and h to ensure practical dimensions for a soup can.