Question

The volume of a cube is decreasing at the rate of 10 m3/hr. How fast is the total surface area decreasing when the surface area is 54 m2?

Answer 1

Explanation:

We have volume of a cube =
`a^3`, where a is the side

Given that
`(dV)/(dt) =10`

`V(a) = a^3 \\S(a) =6a^2`, where S is surface area

Differentiate V wrt t

`(dv)/(dt) =3a^2 (da)/(dt) \\-10 = 3a^2 (da)/(dt)`

To find a:
`S= 54 =6a^2\\a= 3`

Hence we get

`(da)/(dt) =(-10)/(3a^2) \\=(-10)/(27)`

Hence rate of decrease of surface area =
`12a(da)/(dt) =12(-(10)/(27) )(3) = (-40)/(3)`

Decreasing at the rate of 13.33 m^2/hour.