Aright circular cylinder is inscribed in a cone with height h and base radius r. find the largest possible volume of such a cylinder.

 V = (2/9)* π* r²* h

Step-by-step explanation:

Cone dimensions : height  =   h      base radius  = r

In the annex, the angle CDE and DBF ( α ) are iqual since these angles are formed for two paralell lines ( AB and JD ) that intersect the line CB, and the tangent of these angles are:

In triangle CDE  

tan α = CE/ED  where CE = p     CE  = h - l    and ED = × (the unknown radius of the cylinder) then

tan α  = (h - l) ÷ x

And in Triangle DBF

tan α = l ÷ ( r - x ) Therefore

( h - l ) ÷ x  =  l ÷ ( r - x )

Solving  we have      ( h - l ) * ( r - x )  = l*x     ⇒  h*r-h*x -l*r + l*x = l*x

h*r - h*x - l*r = 0         ⇒  l*r  = h*r -h*x      l = h - (h*x) ÷ r

realize x is  base radius of the cylinder

Now the volume of the cylinder is:

V = πr²*l      or   V(x) = π* x² * h - ( π*h* x³ )÷ r

Now taking derivatives

V¨(x) = 2π * x * h - (3π*h*x²) ÷r  

equalizing  to cero        2π *r*h*x - 3π*h*x²  = 0

2r*x - 3x² = 0   ⇒  2r - 3x = 0      x = (2/3)* r  y

l (the height of cylinder is )

l = h - (h/r )*x      ⇒   l = h - (h*x) ÷ r     ⇒   l = [ h (r - x )] ÷r  

l = 1/2 h

And finally the largest possible volume of the cylinder is:

V = π * [ 2/3 r ]² * (1/2) *h

V = (2/9)* π* r²* h

no he doesnt because "-16' is negatiive

step-by-step explanation:

answer: options

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