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# application of differentiation

A right circular cylinder is inscribed in a given right circular cone so that one circular end is on the base of the of the cone and the circumference of the other end on the surface of the cone.Prove that the maximum volume of scuh a cylinder is 4/9 that of the cone.

the two are all did it very well

### 2 Answers

- 六呎將軍Lv 71 decade agoFavorite Answer
Let R and H be the base radius and height of the circular cone respectively, then its volume is πR2H/3.

Then let r and h be the base radius and height of the cylinder respectively. We have, by similar triangle:

(H - h)/H = r/R

1 - h/H = r/R

h/H = 1 - r/R

h = (1 - r/R)H

Thus, in terms of r only, the volume of the cylinder is:

V = πr2h

= πr2(1 - r/R)H

= π(r2 - r3/R)H

dV/dr = πH(2r - 3r2/R)

When dV/dr = 0, 2r - 3r2/R = 0.

Disgarding the ans of r = 0, r = 2R/3

Then h = (1 - 2/3)H = H/3

So max. volume of the cylinder is π(2R/3)2(H/3)

= 4πR2H/27

which is 4/9 of that of the cone.

Source(s): Myself - cipkerLv 51 decade ago
Let h be the fixed height of the cone and r be the fixed base radius of the cone.

Let t be the height of the cylinder and k be the base radius of the cylinder.

Using similar triangles,

k/r = (h -t)/h

t = h(r - k)/r

Volume of cylinder, V= pi k² t = pi k² h(r - k)/r = (pi*h/r)(rk² - k³)

dV/dk = (pi*h/r)(2rk - 3k²)

d²V/dk² = (pi*h/r)(2r - 6k)

Set dV/dk = 0

(pi*h/r)(2rk - 3k²)=0

3k = 2r

k = 2r/3

d²V/dk² │k = 2r/3

=(pi*h/r)[2r - 6*2r/3] =(pi*h/r)(-2r) <0

So V is max. when k = 2r/3.

When k = 2r/3,

Max. V

= (pi*h/r)(rk² - k³)

= (pi*h/r)[r(2r/3)² - (2r/3)³]

=(pi*h)(4r²/27)

=(4/27)*pi*r²h

Volume of cone = (1/3)*pi*r²h

(4/27)*pi*r²h / (1/3)*pi*r²h = 4/9

The maximum volume of scuh a cylinder is 4/9 that of the cone.