Anonymous
Anonymous asked in Science & MathematicsMathematics · 9 months ago

Cylinder Volume Question??? (Picture attached)?

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2 Answers

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  • 9 months ago
    Favorite Answer

    Let H cm be the height of the can when R=4.2 cm, then

    piHR^2=725

    =>

    H=725/(17.64pi)=13.082 cm

    approximately.

    Now, pi[725/(17.64pi)-h](4.2+r)^2=725

    =>

    (725/17.64-pi*h)(17.64+8.4r+r^2)=725

    =>

    h=725(r^2+8.4r)/[17.64pi(r+4.2)^2]

    =>

    (a) h(r)=(725/pi)[1/17.64-1/(r+4.2)^2]

    (b) When the radius increases by 1 cm,

    h(1)=(725/pi)[1/17.64-1/(5.2)^2]=4.547 cm

    h is decreased by 4.547 cm approximately.

    Check:

    the volume of the new can=

    pi(13.082-4.547)(5.2^2)=725.04 ml, valid.

    (c) When the height is reduced by 1 cm,

    (r+4.2)^2=1/(1/17.64-pi/725)

    =>

    r=1/sqr(1/17.64-pi/725)-4.2=0.17 cm

    the radius increases by 0.17 cm approximately.

    Check:

    The volume of the new can=

    pi(13.082-1)(4.2+0.17)^2=724.86 ml, valid.

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  • Mike G
    Lv 7
    9 months ago

    For the original can 725 = πy(4.2)^2 where y = original height in cm

    y = 13.0825 cm

    a) For the new can

    725 = π(13.0825-h)(4.2+r)^2

    13.0825-h = 725/[π(4.2+r)^2]

    h(r) = 13.0825 -725/[π(4.2+r)^2]

    c) (i) h(0.5) = 2.635 cm

    ii) h(1.3) = 5.454 cm

    iii) h(1.5) = 5.980 cm

    d) For this write r as a function of h and then plug in the three values h

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