Question

A foam box's length and width are increasing by 3 and 3 centimeters per minute, respectively, while its height is decreasing by 2 centimeters per minute. At the time when the box's length, width, and height are 1, 2, and 4, respectively, what is the rate of change of the length of the long diagonal of the box

Answer 1

The rate of change of the length of the diagonal
`R` is
`$(dR)/(dt) =(√(21) )/(21) \:cm/min$`

Explanation:

For the box of length
`L`, width
`w`, and height
`h`, the length of the diagonal
`R` is given by

`R^2 = L^2+w^2+h^2`,

Taking the time derivative of both sides we get:

`$(dR^2)/(dt) =(dL^2)/(dt)+(dw^2)/(dt) +(dh^2)/(dt)$`

since for any function
`f(x)`

`(df(t)^2)/(dt) = 2f(t)(df(t))/(dt)`,

our equation becomes

`$2R(dR)/(dt) =2L(dL)/(dt)+2w(dw)/(dt) +2h(dh)/(dt)$`

`$\boxed{(1).\:R(dR)/(dt) =L(dL)/(dt)+w(dw)/(dt) +h(dh)/(dt).}$`

Now, at a certain time when
`L =1, \:w =2,\; h = 4`, the length of the diagonal is

`R^2 = 1+2^2+4^2`

`R = √(21)`,

and since

`(dL)/(dt) = 3cm/min`

`(dw)/(dt) = 3cm/min`

`(dh)/(dt) = -2cm/min`

equation (1) becomes

`$√(21) (dR)/(dt) =(1)(3cm/min)+(2)(3cm/min) +(4)(-2cm/min)$`

`$√(21) (dR)/(dt) =[3+6 +(-8)]\:cm/min$`

`$ (dR)/(dt) =(1)/(√(21) ) \:cm/min$`

`$\boxed{ (dR)/(dt) =(√(21) )/(21) \:cm/min}$`

or
`0.23cm/min.`