Find the maximum volume of a right circular cylinder inscribed in a cone of altitude 12 cm and base radius 4 cm, if the axis of two coincide.

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We have to find the maximum volume of the cylinder.

Now let the height of the required cylinder by equal to h and let its radius be equal to r.

Now the radius of the base of the cone is 4 cm. And the altitude is 12 cm. So the angle made by the line placed along the side is arc tan ( 12 / 4)

For a cylinder to fit into the cone, we require that

arc tan( h/(4-r)= arc tan (12 / 4)

=> h / (4- r) = 12/4 = 3

=> h = 3(4-r)

=> h = 12 - 3r

Now the volume of the cylinder is pi*r^2*h

=> pi*r^2*(12 - 3r)

=> 12*pi*r^2 - 3*pi*r^3

To maximize 12*pi*r^2 - 3*pi*r^3, lets find the derivative

V = 12*pi*r^2 - 3*pi*r^3

V' = 24*pi*r - 9*pi*r^2

Equate V' to 0.

=> 24*pi*r - 9*pi*r^2 = 0

=> 24 - 9*r = 0

=> r = 24/9

=> r = 8/3

Therefore the radius of the cylinder should be 8/3.

The volume then is pi*r^2*h

=> pi*(8/3)^2*4

=> 89.36 cm^3

**The volume of the required cylinder is 89.36 cm^3.**

The volume of the cyllinder is:

V = A base*height

We'll write the volume of the cylinder as a function of x.

The area of the base of the cyllinder is:

A = pi*r^2

We'll determine r using proportions:

R/H = r/(12-x)

R = 4, H = 12, h is the height of the cylinder and r is the radius of the cylinder.

4/12 = r/(12-x)

1/3 = r/(12-x)

We'll cross multiply and we'll get:

3r = 12 - x

r = (12-x)/3

The volume of the cylinder is:

V(x) = pi*x*(12-x)^2/9

V(x) = (pi/9)*(144x - 24x^2 + x^3)

The volume is maximum for the critical value of x that cancels the first derivative of V(x).

We'll determine V'(x):

V'(x) = (pi/9)*(144 - 48x + 3x^2)

V'(x) = 0

144 - 48x + 3x^2 = 0

We'll divide by 3:

x^2 - 16x + 48 = 0

We'll apply quadratic formula:

x1 = [16+sqrt(256 - 192)]/2

x1 = (16+8)/2

x1 = 12 cm

x2 = (16-8)/2

x2 = 4 cm

For x = 12 cm, the volume of the cylinder is maximum.

The let the height of the cone be h and radius r inscribed in the cone.

The vertical semiangle of the cone with axis , x = arc tan (4/2*12) = arc tan (1/6) .

x = arc tan (1/6)

Therefore the radius of the cylinder r = (12-h) tan x = (1-h)/6, where h is the height of cone.

Therefore the volume of the cylinder v(h) = pi r^2*h = pi*{(12-h)^2/6^2}*h.

To find maximum v(x) , we find x = c , for which v'(c) = 0 and v"(c) < 0.

v'(x) = p^2/36 { (12-h)^2*h}'

v'(x) = p^2/36{ 144h - 24h^2+h^3}'

v'(x) = p^2/36{144-48h+3h^2}

v'(x) = 0 gives 144-48h+3h^2 = 0. Or

Divide by 3:

48-16h+h^2 = 0.

(h-12)(h-4) = 0

So h = 4, or h= 12.

v"(x) = pi/36{ -48 +6h} < 0 for x= 4 as -48+6*4 = -24.

Therefore the volume of the inscribed cylinder is maximum for x = 4 and the maximum volume is :

v(4) = (pi/36){ (12-4)^2* 4} = (pi/36)(64*4)= 64pi/9 = 22.34. sq units.

The height of the cone = 4 and the radius of the cone = (12-4)/6 = 4/3.

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