Calculus Related Rates (A mathematical expanding box)?

I am confused on where to even start on this question, Id greatly appreciate it if I could get guided step by step on this question, thank you very much.

A mathematical expanding box has sides of length l=2cm, w=3cm and h=4cm at t=0. The length is expanding at a rate of 0.4 centimeters per second, the width is expanding at a rate of 0.3 centimeters per second and the height is expanding at a rate of 0.2 centimeters per second. How fast is the volume of the box increasing at the moment when it is a cube? 

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  • 2 months ago
    Favorite Answer

    l = 2, I’(t)= 0.4

    w = 3, w’(t) = 0.3

    h = 4, h’(t) = 0.2

    dl/dt = 0.4

    l = 0.4t + 2

    dw/dt = 0.3

    w = 0.3t + 3

    dh/dt = 0.2

    h = 0.2t + 4

    Volume V

    V = lwh

    V = ( 0.4t + 2 )( 0.3t + 3 ) ( 0.2t + 4 )

    V = 0.024t³ + 0.84t² + 8.4t + 24

    cube:

    ( 0.4t + 2 )( 0.3t + 3 ) ( 0.2t + 4 ) = ( 0.4t + 2 )³

    t = -8, 

    t = -5,

    t = 10

    ignore negative times

    t = 10

    V = 0.024t³ + 0.84t² + 8.4t + 24

    V’(t) = 0.072t² + 1.68t + 8.4

    V’(10) = 0.072*10² + 1.68t + 8.4

    V’(10) = 32.4 cm³/s

    ——————————

  • ted s
    Lv 7
    2 months ago

    V = l w h...find dV/dt  when l = w = h and  you know dw/dyt = 0.3 , dl/dt = 0.4 ,

    dh / dt = 0.2  & l = 2 + 0.4 t , w = 3 + 0.3 t , h = 4 + 0.2 t 

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