Answer:
The maximum volume of inscribed cylinder when its radius is r = 10 \3 u n i t s and its height is 2 units.
Explanation:
The formula to find the volume of inscribed cylinder is: V = (πr^2).y
let 'y' is the height of the cylinder and 'r' be the radius of the cylinder.
Radius of the inscribed cylinder is r = 5-x.
and the value of y is y = (5.5)/5 * x, it is find by the formula of slope because we know the slope of the slanted side of the cone. We go up 5.5 units for every 5 units we move to the right.
Now,
V = π(5-x)^2 × (5.5/5) . x
V = (5.5/5 . π) × (25x + x^3 - 10x^2)
And now to find the maximum volume derivative of V with respect to 'x' will be equal to zero like: dV/dx = 0
dV/dx = (5.5/5 . π) × (25 + 3x^2 - 20x)
(5.5/5 . π) × (25 + 3x^2 - 20x) = 0
25 + 3x^2 - 20x = 0
Write the equation in sequence and then use the factorization to find x
3x^2 - 20x + 25 = 0
3x^2 - 15x - 5x + 25 = 0
3x(x-5) -5(x-5) = 0
(x-5) (3x-5) = 0
x = 5/3 and x = 5
So, we have two values of 'x'. But the value 5/3 gives us the point of maximum by graphing V with 'x'. Because when the value is between 0 to 5 then the left edge of the inscribed cylinder stays properly within the cone.
Hence, at x = 5/3,
y = (5.5)/5 * x = (5.5)/5 * 5/3
y = 1.833 ≈ 2.
and r = 5 - x = 5 - 5/3 = 10/3.