Math - Calculus I

Optimization Problem:

Find the dimensions of the right circular cylinder of greatest volume inscribed in a right circular cone of radius 10" and height 24"

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  1. Draw a side view of the situation
    (The cone will look like an isosceles triangle with a rectangle (the cylinder) sitting on its base and touching the sides)

    let the radius of the cylinder be r and let the height of the cylinder be h
    Look at the small right-angled triangle at the right
    we can set up a ratio because of similar triangles
    h/(10-r) = 24/10
    10h = 240-24r
    h = 24 - 12r/5

    V = πr^2 h
    = πr^2 (24 - 12r/5)
    =24πr^2 - (12π/5) r^3

    dV/dr = 48πr - (36π/5)r2
    = 0 for a max/min

    48πr = (36π/5)r^2
    ÷ by 12π
    4r = 3π/5 r^2
    divide by r, since r ≠ 0, it would give a minimum volume
    4 = 3r/5
    20 = 3r
    r = 20/3
    then h = 24 - (12/5)(20/3) = 8

    the cylinder has a radius of 20/3 and a height of 8

    check:
    with those dimensions, V = 1117.01
    let r = 19/3 (a little less) , then h = 44/5 , V = 1108.9 , a bit less
    let r = 21/3 ( a little more), then h = 36/5 , V = 1108.35 (again, a little less)

    My answer looks reasonable

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