Posted by **REALLY NEED HELP!!!!** on Friday, October 29, 2010 at 6:00pm.

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?

- Math: Calculus -
**bobpursley**, Friday, October 29, 2010 at 6:25pm
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.

- Math: Calculus -
**REALLY NEED HELP!!!!**, Friday, October 29, 2010 at 6:36pm
I manage to get to that point but maximizing it is the problem.

- Math: Calculus -
**Damon**, Friday, October 29, 2010 at 6:36pm
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 =

- error -
**Damon**, Friday, October 29, 2010 at 6:53pm
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

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