Question

The surface area of a cube is increasing at a rate of 15 square meters per hour. At a certain instant, the surface area is 24 square meters. What is the rate of change of the volume of the cube at that instant (in cubic meters per hour)?

Answer 1

Answer:

The volume of cube is increasing at a rate 7.5 cubic meter per hour.  

Step-by-step explanation:

We are given the following in the question:

Surface area of cube = 24 square meters.

Let l be the edge of cube.

Surface area of cube =

$6l^2 = 24\\l^2 = 4\\l = 2$

Thus, at that instant the edge of cube is 2 meters.

$\dfrac{dS}{dt} = 15\text{ square meters per hour}\\\\\dfrac{d(6l^2)}{dt} = 15\\\\12l\dfrac{dl}{dt} = 15\\\\\dfrac{dl}{dt} = \dfrac{15}{12\times 2} = \dfrac{15}{24}$

We have to find the rate of change in volume.

Volume of cube =

$l^3$

Rate of change of volume =

$\dfrac{dV}{dt} = \dfrac{d(l^3)}{dt} = 3l^2\dfrac{dl}{dt}\\\\\dfrac{dV}{dt} = 3(2)^2\times \dfrac{15}{24} \\\\\dfrac{dV}{dt} = 7.5\text{ cubic meter per hour}$

Thus, the volume of cube is increasing at a rate 7.5 cubic meter per hour.