Polynomial functions problem thinking question help?????

i do not get how to solve this question

A packaging company has been asked to make cylindrical cans that can hold a volume of 16π cm3. The company owns a machine that only makes cans that have a height which is the square of 3 less than the radius. To achieve the desired volume, what should the dimensions of the finished can be?

2 Answers

  • Anonymous
    5 years ago
    Favorite Answer

    radius = r

    height = (r-3)^2

    Volume = πr^2(h) = πr^2(r-3)^2 = 16π

    Writ3 that as

    r^2(r-3)^2 - 16 = 0

    You don't want to have to solve a 4th degree equation. Fortunately, you don't have to. You have the difference of two squares, so you can easily factor that as

    (r(r-3) - 4)(r(r-3) + 4) = 0 so one of those terms must = 0

    Look at both of them to see all possible solutions.

    r^2 -3r - 4 = 0

    (r-4)(r+1) = 0 ==> r=4 is only sensible (positive) value.

    r^2 - 3r + 4 = 0 has no real solutions.

    Answer: 4cm radius, (4-3)^1 = 1 cm high.

    Volume is π4^2(1) = 16π

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