A container manufacturing company was contracted to design and manufacture cylindrical cans for fruit juice. The volume of each can needs to be 0.946 liters. In order to minimize production costs, the company wishes to design cans that require the smallest amount of material possible. (This means that you need to design a can with volume=.946 liters and the smallest possible surface area. Explain the relationship between the surface area and the amount of material.)

The volume of the cylinder is the amount of fruit juice it can contain.

The relationship between the volume and the surface area is:

[tex]\mathbf{A = \pi (\sqrt[3]{\frac{V}{2\pi}})^2 + \frac{0.946}{(\sqrt[3]{\frac{V}{2\pi}})}}[/tex]

The given parameter is:

[tex]\mathbf{V = 0.946}[/tex]

The volume of a cylinder is calculated as:

[tex]\mathbf{V = \pi r^2 h}[/tex]

Make h the subject

[tex]\mathbf{h = \frac{V}{ \pi r^2}}[/tex]

The surface area (A) of a cylinder is:

[tex]\mathbf{A = \pi r^2 + \pi rh}[/tex]

Substitute [tex]\mathbf{h = \frac{V}{ \pi r^2}}[/tex]

[tex]\mathbf{A = \pi r^2 + \pi r \times \frac{V}{\pi r^2}}[/tex]

[tex]\mathbf{A = \pi r^2 + \frac{V}{r}}[/tex]

Differentiate

[tex]\mathbf{A' = 2\pi r - Vr^{-2}}[/tex]

Set to 0

[tex]\mathbf{2\pi r - Vr^{-2} = 0}[/tex]

Rewrite as:

[tex]\mathbf{ 2\pi r = Vr^{-2}}[/tex]

Multiply through by r^2

[tex]\mathbf{ 2\pi r^3 = V}[/tex]

Solve for r

[tex]\mathbf{r = \sqrt[3]{\frac{V}{2\pi}}}[/tex]

[tex]\mathbf{A = \pi r^2 + \frac{0.946}{r}}[/tex]

So, we have:

[tex]\mathbf{A = \pi (\sqrt[3]{\frac{V}{2\pi}})^2 + \frac{0.946}{(\sqrt[3]{\frac{V}{2\pi}})}}[/tex]

Hence, the relationship between the volume and the surface area is:

[tex]\mathbf{A = \pi (\sqrt[3]{\frac{V}{2\pi}})^2 + \frac{0.946}{(\sqrt[3]{\frac{V}{2\pi}})}}[/tex]

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