Question

Help please:))

2. When shipping ice cream, melting is understandably a big concern. You will notice that ice cream is not generally packaged in a cube-shaped container. A standard container of ice cream contains 1 L, or 1000 cm3 of ice cream,

a. What would be the optimal dimensions (radius and height) to minimize surface area?

b. What would the surface area be?

C. Suggest at least two reasons why this is different from the ice cream packaging that you see in the stores.​

Answer 1

a) Because this asks about the radius and height, I assume that we are talking about a cylinder shape.

Remember that for a cylinder of radius R and height H the volume is:

V = pi*R^2*H

And the surface will be:

S = 2*pi*R*H + pi*R^2

where pi = 3.14

Here we know that the volume is 1000cm^3, then:

1000cm^3 = pi*R^2*H

We can rewrite this as:

(1000cm^3)/pi = R^2*H

Now we can isolate H to get:

H = (1000cm^3)/(pi*R^2)

Replacing that in the surface equation, we get:

S = 2*pi*R*H + pi*R^2

S = 2*pi*R*(1000cm^3)/(pi*R^2) + pi*R^2

S = 2*(1000cm^3)/R + pi*R^2

So we want to minimize this.

Then we need to find the zeros of S'

S' = dS/dR = -(2000cm^3)/R^2 + 2*pi*R = 0

So we want to find R such that:

2*pi*R = (2000cm^3)/R^2

2*pi*R^3 = 2000cm^3

R^3 = (2000cm^3/2*3.14)

R = ∛(2000cm^3/2*3.14) = 6.83 cm

The radius that minimizes the surface is R = 6.83 cm

With the equation:

H = (1000cm^3)/(pi*R^2)

We can find the height:

H = (1000cm^3)/(3.14*(6.83 cm)^2) = 6.83 cm

(so the height is equal to the radius)

b) The surface equation is:

S = 2*pi*R*H + pi*R^2

replacing the values of H and R we get:

S = 2*3.14*(6.83 cm)*(6.83 cm) + 3.14*(6.83 cm)^2 = 439.43 cm^2

c) Because if we pack cylinders, there is a lot of space between the cylinders, so when you store it, there will be a lot of space that is not used and that can't be used for other things.

Similarly for transport problems, for that dead space, you would need more trucks to transport your ice cream packages.