# A right circular cylinder is inscribed in a sphere of diameter 8 cms. If the cylinder is open at both ends, find the largest possible surface area of the cylinder.

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The surface area of a cylinder, without the ends, is `LA=2pirh`

Since the cylinder is a right circular cylinder, we can draw a right triangle. Select a point on one of the ends. Draw a diameter from this point. From the point draw a segment perpendicular to the diameter to the other end of the cylinder.

The right triangle will have sides 2r,h, and 8. (2r since it is a diameter of the circle; h is the height of the cylinder; 8 because connecting the other end of the diameter to the far point is a diamter of the sphere.)

Then by the Pythagorean theorem we have `(2r)^2+h^2=8^2` or `h=sqrt(64-4r^2)`

Rewriting our formula for the lateral area we get:

`LA=2pirsqrt(64-4r^2)`

To maximize, we take the first derivative to find the critical points.

`(dLA)/(dr)=2pi(64-4r^2)^(1/2)+2pir(1/2)(64-4r^2)^(-1/2)(-8r)`

`=(64-4r^2)^(-1/2)(2pi(64-4r^2)-8pir^2)`

`=(128pi-16pir^2)/sqrt(64-4r^2)`

For a fraction to be zero, we need the numerator to be zero when the denominator is nonzero.

`128pi-16pir^2=0`

`r^2=8`

`r=2sqrt(2)`

then `h=sqrt(64-4(8))=4sqrt(2)`

Then the lateral area (the surface area not including the ends) is :

`LA=2pi(2sqrt(2))(4sqrt(2))=32pi`