Respuesta :
Answer: 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
The largest possible volume of such a right circular cylinder is gotten as; V_max = 4πr²h/27
What is the largest volume of the Cylinder?
The main equation to maximize the volume of the cylinder is;
V = πx²y
Using common ratios of similar triangles to get:
Ht of triangle above cylinder/Base of Triangle above cylinder = Ht of full triangle/ Base of full triangle
Thus;
(h - y)/2x = h/2r
Cross multiply to get;
h – y = hx/r
y = h – hx/r
y = h(1 – x/r)
Plug in this expression of y into our Volume equation:
V(x) = πx²h(1 – x/r)
V(x) = πh(x² - x³/r)
To find the maximum volume, we need to first get the derivative and solve for when the derivative equals zero. Thus;
V’(x) = πh(2x - 3x²/r) = 0
(2x - 3x²/r) = 0
x( 2r – 3x) = 0
Thus;
x = 0 or x = 2r/3
we will use only x = 2r/3
Put 2r/3 for x into our volume equation to get;
V_max = πh ((2r/3)² – (2r/3)³/r)
V_max = πh (4r²/9 - 8r²/27)
V_max = 4πr²h/27
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