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# 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?

### 2 Answers

- King LeoLv 72 months agoFavorite 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

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- ted sLv 72 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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Very helpful, thanks a ton