Inscribed Cylinder?

Question

A cylinder is inscribed in a right circular cone of height 8 and radius (at the base) equal to 4 . What are the dimensions of such a cylinder which has maximum volume? What is the height and radius:

Here is what I have:

plotting the cone on a graph gives me points:
(0,8) - tip of cone
(4,0) - radius of cone

from this, i figure out the formula of the line that is tangent to the point that the tip of the side of cylinder touches:

y= mx + b = -2x+8

The formula for the volume of a cylinder is V=pir^2h

the formula for the line can be changed in terms of r anfd h (radius and height respectively) to be:

h = -2r+8

Plugging this into the eqn: V=pi*r^2*(-2r+8)

But then... I get stuck - if i differentiate, I still have two variables dV/dx and r.

Is the process wrong? Have I skipped something?

Answer 1

You actually have dV/dr!
Since you need the maximum, set the derivative to zero then you can get the r. Choose r between 0 & 4.

Answer 2

Choose a 2-D coordinate plane such that the intersection point to the right hand side is on the line of x/4 + y/8 = 1, which can be simplified into y = 8-2x
vol = pi x^2*y = pi x^2(8-2x)
Solve d vol/dx = 0 for x,
x = 8/3
Therefore, the maximum volume of the cylinder is
v = pi (8/3)^2(8-16/3) = 59.57 units^3
-------------
Your approach works too. You just need to solve dV/dr = 0 for r.

Answer 3

don't worry about including pi in the equation at first because all of the possible volumes will have that in common. then don't worry about a maximum volume of the cylinder but rather a minimum volume of the inscribed cone; using those thoughts, put the problem to paper and graph the cone.
the cone will be triangular and the cylinder rectangluar. you want the largest possible area of the the cylinder or smallest possible area of the cone
you sound like you'll be able to figure it out from there

Answer 4

20^2 = r^2 + (h/2)^2 using Pythagoras' theorem on a corner of the cylinder. 2*sqrt(4 hundred - r^2) = h A) V = pi*(r^2)*h = pi*(r^2)*2*sqrt(4 hundred - r^2) B) dV/dr = 4pi*r*sqrt(4 hundred - r^2) - 2pi*(r^3)/sqrt(4 hundred - r^2) = 0 0 = 4pi*r*(4 hundred - r^2) - 2pi*r^3 0 = 2*(4 hundred - r^2) - r^2 r^2 = 800/3, replace this to V to get V = 19347 contraptions^3 needless to say the minimum volume is 0 if r=20.

Answer 5

You differentiate with respect to r, not x, so
V = πr^2*(8 - 2r) = 8πr^2 - 2πr^3
dV/dr = 16πr - 6πr^2 = 0
r(3r - 8) = 0
r = 0 (minimum)
r = 8/3
h = 8 - 2*8/3
h = (24 - 16)/3 = 8/3
V = π(8/3)^@(8/3) = π(8/3)^3
V ≈ 59.57390