Make a cross-section diagram showing a rectangle within a triangle, or if you are artistically inclined, make a drawing of the cylinder within the cone.
Let the radius of the cylinder be r,
let its height be h
by similar triangles,
h/(2-r) = 5.5/2
2h = 11 - 5.5r
h = (11-5.5r)/2
volume of cylinder
= πr^2 h
= πr^2 (11 - 5.5r)/2
= (11/2)πr^2 - (5.5/2π r^3
d(volume)/dr = 11πr - (16.5/2)π r^2
= 0 for a max of volume
11πr - 8.25πr^2 = 0
divide out the π and factor out an r
r(11 - 8.25r) = 0
r = 0 or r = 11/8.25
clearly r = 0 would give a "minimum" so
r = 11/8.25 or 4/3
the h = (11 - 5.5(4/3) )/2 = 11/6
A) to obtain a maximum volume,
the radius is 4/3 and the height is 11/6
B) and C) answered in A)
check my arithmetic
A cylinder is inscribed in a right circular cone of height 5.5 and radius (at the base) equal to 2 .
A) What are the dimensions of such a cylinder which has maximum volume?
B) What is the radius?
C) What is the height?
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