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
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