Math: Calculus
posted by REALLY NEED HELP!!!! on .
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?

Well, the upper radius= lower radius(1h/3.5)
Volume cylinder=PIr^2*h
volumecylinder= PI (7^2(1h/3.5)^2 h)
I assume you can maximize that. 
I manage to get to that point but maximizing it is the problem.

inscribe rectangle of base r and height h in triangle of height 3.5 and base 7
similar triangles top and bottom
(3.5h)/r = h/(7r)
hr = (3.5h)(7r) = 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(96047203) ] / 98
h = 1 +/ 1/2
h = 1.5 or 0.5
try both of those to see if one is bigger
dv/dh = 
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(31362352) ] / 24
h = 2.33 +/ 1.17
h = 3.5 or .56
3.5 is zero volume so try .56