PhysicsinCalifornia
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HW problem--NEED HELP!
This is an optimization problem with differential calc. I really need help with this
Here's the prob:
A right circular cone is being inscribed in a sphere of radius 3cm.
Find
a) the dimensions (base radius, and height) of the right circular cone with the largest volume
b) the cone's volume
Here's what I got::
Volume for sphere is
V_s = \frac{4}{3}\pi r^3
and the volume for the cone is
V_c = \frac{1}{3}\pi r^2 h
(obviously)
Now i got that h = 3+x
*Note that I cannot draw a pic, so it's hard to describe
Also, x = \sqrt{9 - r^2} using pythagorean's theorem
so the height would equal the radius of the sphere, 3 cm, plus x
Therefore
V(r) = \frac{1}{3} \pi r^2 (3 + \sqrt{9- r^2}) = \pi r^2 + \frac{\pi r^2 \sqrt{9 - r^2}}{3}
leaving everything in terms of r because we want the largest volume
How do I work it from here?
Thanks for your help in advance
This is an optimization problem with differential calc. I really need help with this
Here's the prob:
A right circular cone is being inscribed in a sphere of radius 3cm.
Find
a) the dimensions (base radius, and height) of the right circular cone with the largest volume
b) the cone's volume
Here's what I got::
Volume for sphere is
V_s = \frac{4}{3}\pi r^3
and the volume for the cone is
V_c = \frac{1}{3}\pi r^2 h
(obviously)
Now i got that h = 3+x
*Note that I cannot draw a pic, so it's hard to describe
Also, x = \sqrt{9 - r^2} using pythagorean's theorem
so the height would equal the radius of the sphere, 3 cm, plus x
Therefore
V(r) = \frac{1}{3} \pi r^2 (3 + \sqrt{9- r^2}) = \pi r^2 + \frac{\pi r^2 \sqrt{9 - r^2}}{3}
leaving everything in terms of r because we want the largest volume
How do I work it from here?
Thanks for your help in advance
Last edited: