Circular cone has slant height of 8 cm. What are dimensions of cone to product largest volume?
We know that the volume of the cone is given by :
V = (1/3)*pi * r^2 * h such that r is the radius and h is the height
But we know that the slant is given by s = sqrt( h^2 + r^2)
==> s^2 = h^2 + r^2
==> r^2 = s^2 - h^2
Given that the slant s = 8
==> r^2 = 64- h^2
Now we will substitute into the volume,
==> V = (1/3)*pi* h * ( 64-h^2)
==> V = (1/3)pi*( 64h - h^3)
==> (64pi/3)h - pi/3 * h^3
Now we know that the maximum point if the derivatives zero.
==> v' = 64pi/3 - pi*h^2 = 0
==> h^2 = 64pi/3pi = 64/3
==> h= 8/sqrt3 = 4.62 cm
Now we will find r.
==> r^2 = s^2 - h^2 = 64 - 64/3 = 128/3
==> r= sqrt(128/3) = 8sqrt(2/3)= 6.532 cm
Then the dimensions of the cone are : radius = 6.532 cm and the height h= 4.62 cm
The volume of the cone is (1/3)*pi*r^2*h.
Let the radius be r and the height be h. According to Pythagorean Theorem, r^2 + h^2 = 8^2
=> r^2 = 64 - h^2
This gives the volume as V = (1/3)*pi*(64 - h^2)*h
=> V = (1/3)*pi*(64h - h^3)
To maximize volume we have to find dV/dh and solve for h using dV/dh = 0
dV/dh = (1/3)*pi*(64 - 3h^2)
(1/3)*pi*(64 - 3h^2) = 0
=> 3h^2 = 64
=> h^2 = 64/3
=> h = 8/sqrt 3
r = sqrt(64 - 64/3) = sqrt(2*64/3)
For the largest volume the radius should be 8*sqrt(2/3) and the height should be 8/(sqrt 3)