A cylinder is inscribed in a right circular cone of height 4.5 and radius (at the base) equal to 5. What are the dimensions of such a cylinder which has maximum volume

The maximum volume of inscribed cylinder  when its radius is r  =  10 \3 units  and its height is  1.5 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 = (4.5)/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 4.5 units for every 5 units we move to the right.

Now,

V = π(5-x)^2 × (4.5/5) . x

V = (4.5/5 . 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 = (4.5/5 . π) × (25 + 3x^2 - 20x)

(4.5/5 . 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 = (4.5)/5 * x =  (4.5)/5 * 5/3

y = 1.5 ≈ 2.

and r = 5 - x = 5 - 5/3 = 10/3.

The maximum volume of inscribed cylinder  when its radius is r  =  10 \3 units  and its height is  1.5 units.

pe = weightxheight = 2000x1000=2,000,000joules

frequency or wavelenth of em radiation in a vacuum