HELP ASAP-a cylindrical wire frame for a wastebasket has a wire circle for the top and bottom and 6 straight?

Wastebasket

D asks:

wire rods that hold the top and bottom circles together. The total amount of wire used is 4m. What radius and height will maximize the volume of the cylindrical frame? Round your answer to the nearest thousandth of a metre.

this is an optimization problem.

Best answer:

Answer by Perplexed
Use the radius r of the top/bottom as your variable. Call h the height of the cylinder. Therefore,

2*pi*r + 2*pi*r + 6*h = 4, which will let you express h as a function of r.

Now V (volume) = pi*r*r*h. Simply express h as a function of r (using the equation above), differentiate V with respect to r, equate to zero and solve for r (you might get more than one solution though, check them all).

2 thoughts on “HELP ASAP-a cylindrical wire frame for a wastebasket has a wire circle for the top and bottom and 6 straight?

  1. V = pi*h*r^2
    but
    6h+2*(2pi*r) = 4

    h = 2/3*(1-pi*r)

    V = pi * 2/3 *(1-pi*r)*r^2
    = 2/3pi*(r^2 -pi * r^3)
    dv/dr = 2/3pi*(2r-3pi*r^2)
    for dv/dr = 0 to get max volume
    2r-3pir^2 = 0
    r(2-3pi*r) =0
    which get r = 0 not acceptable it gives min volume
    or 2-3pi*r = 0
    r= 2/(3pi)
    r = 0.212 meter

    which get h = 2/9 = 0.222 meter

    and V = 0.0314 cubic m

  2. Let the radius = r
    the height of the cylinder = h .
    Therefore,
    2*pi*r + 2*pi*r + 6*h = 4, ……………… [1]
    Differentiate w.r.t. r, we get,
    2pi +2pi + 6dh/dr = 0
    6.dh/dr = 4pi
    dh/dr = 4pi/6
    dh = (4pi/6).dr
    h = (4pi/6).r
    From [1], we get
    2*pi*r + 2*pi*r + 6*(4pi/6)r = 4
    8pir = 4
    r = 1/2.pi = 0,1592 m …………………. Answer
    h = 1/3 m …………………. Answer

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