Question

The volume of a cube is decreasing at a rate of 0.5 in3 {min. How fast is the surface area of the cube decreasing when the length of an edge is 0.6 inches

Answer 1

The surface area of the cube decreasing at a rate of 3.33 square inches/min.

Explanation:

It is given that, the volume of a cube is decreasing at a rate of 0.5 cubic inches/min.

`(dV)/(dt)=-0.5\text{ in}^3/min` ...(1)

We know that, volume of a cube is

`V=a^3`

Differentiate w.r.t. t.

`(dV)/(dt)=3a^2(da)/(dt)`

`-0.5=3a^2(da)/(dt)`

`-(0.5)/(3a^2)=(da)/(dt)` ...(2)

Surface area of the cube is

`A=6a^2`

Differentiate w.r.t. t.

`(dA)/(dt)=12a(da)/(dt)`

`(dA)/(dt)=12a\left(-(0.5)/(3a^2)\right)`
`[Using (2)]`

`(dA)/(dt)=-(2)/(a)`

We need to find how fast is the surface area of the cube decreasing when the length of an edge is 0.6 inches.

Substitute a=0.6 in the above equation.

`(dA)/(dt)=-(2)/(0.6)`

`(dA)/(dt)=-3.33`

Therefore, the surface area of the cube decreasing at a rate of 3.33 square inches/min.