Question

At a certain instant the edge of a cube is 7 inches long and the volume is increasing at the rate of 3 cubic inches per minute. How fast is the surface area of the cube increasing

Answer 1

The surface area increases at 12/7 square inches per minute

Explanation:

Given

`l = 7` --- side length

`(dV)/(dt) = 3in^3/min` --- Rate at which volume increases

Required

The rate at
`which` the
`surface` area
`increases`

The volume of a cube is:

`V = l^3`

Differentiate with respect to time

`(dV)/(dt) = 3l^2 \cdot (dl)/(dt)`

When
`l = 7` and
`(dV)/(dt) = 3in^3/min`

The expression becomes:

`3 = 3 * 7^2 \cdot (dl)/(dt)`

Divide both sides by 3

`1 = 7^2 \cdot (dl)/(dt)`

`1 = 49 \cdot (dl)/(dt)`

Divide both sides by 49

`(dl)/(dt) = (1)/(49)`

The surface area is calculated as:

`A= 6l^2`

Differentiate with respect to time

`(dA)/(dt) = 12l(dl)/(dt)`

Substitute:
`(dl)/(dt) = (1)/(49)` and
`l = 7`

`(dA)/(dt) = 12 * 7 *(1)/(49)`

`(dA)/(dt) = 12 *(1)/(7)`

`(dA)/(dt) = (12)/(7)`

The surface area increases at 12/7 square inches per minute