Question

A cylinder is inscribed in a right circular cone of height 3.5 and radius (at the base) equal to 7. What are the dimensions of such a cylinder which has maximum volume?

Answer 1

Well, the upper radius= lower radius(1-h/3.5)

Volume cylinder=PIr^2*h

volumecylinder= PI (7^2(1-h/3.5)^2 h)
I assume you can maximize that.

Answer 2

I manage to get to that point but maximizing it is the problem.

Answer 3

inscribe rectangle of base r and height h in triangle of height 3.5 and base 7
similar triangles top and bottom
(3.5-h)/r = h/(7-r)
hr = (3.5-h)(7-r) = 24.5 - 3.5 r - 7 h + h r
so
7 h + 3.5 r = 24.5
r = 7 - 3.5 h
(there are easier ways of figuring that out)

now the cylinder volume
v = pi r^2 h
v = pi (49 - 49 h + 12.25 h^2)h
v = pi (49 h -49 h^2 +12.25 h^3)
dv/dh = pi (49 - 98 h + 36.75 h^2)
= 0 for max or min
h = [ 98 +/-sqrt(9604-7203) ] / 98
h = 1 +/- 1/2
h = 1.5 or 0.5
try both of those to see if one is bigger

dv/dh =

Answer 4

7 h + 3.5 r = 24.5
r = 7 - 2 h
(there are easier ways of figuring that out)

now the cylinder volume
v = pi r^2 h
v = pi (49 - 28 h + 4 h^2)h
v = pi (49 h -28 h^2 + 4 h^3)
dv/dh = pi (49 - 56 h + 12 h^2)
= 0 for max or min
h = [56 +/-sqrt(3136-2352) ] / 24
h = 2.33 +/- 1.17
h = 3.5 or .56
3.5 is zero volume so try .56