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This can be done by using normal ordianry differentiation techniques.
Therotically the there are two extreme dimensions for a cylinder to be inscribed in such cone. First one, cylinder can have the maximum height of 3 and zero radius. Second one, the cylinder can have zero height and the maximum radius of 5.5. So ideally and cylinder will have the dimensions in those range, height from (0 to 3) while the radius is (5.5 to 0). So let r be the radius of our cylinder and h be the height. So,
0<r<5.5 and 0<h<3
To satisfy this requirement we can have a number of cylinders and according to the dimensions the volume will change. First we will develop an expression for the volume, V using r and h.
`V = pir^2*h`
But since the cylinder is inscribed inside the cone, we can find a relationship for r and h, by using simple geometry and ratios. You will see,
`h / (5.5-r) = 3/5.5`
therefore we can derive an expression for h using r, (5.5 = 11/2)
`h = 3 - 6/11*r`
We can substitutie this in above equation for V,
`V = pi*r^2*3*(1-2/11*r)`
`V = 3*pi*(r^2-2/11*r^3)`
differentiate this wrt to r,
`(dV)/(dr) = 3pi*(2r -6/11*r^2)`
`(dV)/(dr) = 6pi*(r -3/11*r^2)`
for maxima or minima,
`(dV)/(dr) = 0`
`r -3/11*r^2 = 0`
the answers are,
`r = 0 ` `1 - 3*r/11 = 0`
But radius is never equal to zero. Therefore, you get,
`1 -3*r/11 = 0`
`r = 11/3 = 3.6667 < 5.5`
Calculate the second derivative, in order to find whether this is a maximum or minimum point.
`(d^2V)/(dr^2) = 6pi(1 - 6*r/11)`
Substitute r = 11/3 and check for the sign of second derivative.
`(d^2V)/(dr^2) = 6pi(1 - 6*11/(11*3))`
`(d^2V)/(dr^2) = 6pi(1 - 2)`
`(d^2V)/(dr^2) = -6pi`
The sing of teh second derivative is negative, which means the point is a maximum point. Therefore at r = 11/3 the cylinder will have the maximum volume.
The height at that time can be found like this,
`h = 3 - 6/11*r`
`h = 3 - 6/11*(11/3)`
`h = 3 - 2 = 1`
Therefore the hreight at that point is 1.
The dimensions of the cylinder with maximum volume is,
r = 11/3 and h = 1.
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