# Find the values of r which minimizes the surface area.TIn cans come in various shapes and sizes, but what factors influence their design? In particular, is minimising the area of tin used to make...

Find the values of r which minimizes the surface area.

TIn cans come in various shapes and sizes, but what factors influence their design? In particular, is minimising the area of tin used to make a can an important factor?

Suppose a manufacture wishes to enclose a fixed volume, V, using a cylindrical can. The height of the cylinders denoted by, h, and the radius of the cylinder can section by r.

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We know that the surface area of the cylinder is given by :

SA = 2pir^2 + 2pi*r*h

But we know that the volume is :

v = r^2 * pi * h

Given that v is a fixed value.

Then we will write h as a function of r.

==> h = v/r^2 * pi

Now we will substitute into SA.

==> SA = 2pi*r^2 + 2pi*r *(v/r^2 pi)

==> SA = 2pi*r^2 + 2v/ r

Now we will differentiate:

==> (SA)' = 4pir - 2v/r^2

Now we will find critical values.

==> 4pi*r - 2v/r^2 = 0

==> 2v/r^2 = 4pi*r

==> Multiply by r^2

==> 2v = 4pi*r^3

==> r^3 = v/2pi

==> r = (v/2pi)^1/3

**Then the value of r that given minimum surface area is r= (v/2pi)^1/3**