Draw a side view of the situation
(The cone will look like an isosceles triangle with a rectangle (the cylinder) sitting on its base and touching the sides)
let the radius of the cylinder be r and let the height of the cylinder be h
Look at the small right-angled triangle at the right
we can set up a ratio because of similar triangles
h/(10-r) = 24/10
10h = 240-24r
h = 24 - 12r/5
V = πr^2 h
= πr^2 (24 - 12r/5)
=24πr^2 - (12π/5) r^3
dV/dr = 48πr - (36π/5)r2
= 0 for a max/min
48πr = (36π/5)r^2
÷ by 12π
4r = 3π/5 r^2
divide by r, since r ≠ 0, it would give a minimum volume
4 = 3r/5
20 = 3r
r = 20/3
then h = 24 - (12/5)(20/3) = 8
the cylinder has a radius of 20/3 and a height of 8
check:
with those dimensions, V = 1117.01
let r = 19/3 (a little less) , then h = 44/5 , V = 1108.9 , a bit less
let r = 21/3 ( a little more), then h = 36/5 , V = 1108.35 (again, a little less)
My answer looks reasonable
Optimization Problem:
Find the dimensions of the right circular cylinder of greatest volume inscribed in a right circular cone of radius 10" and height 24"
1 answer
0 answers