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)?

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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 =

[tex]6l^2 = 24\\l^2 = 4\\l = 2[/tex]

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

[tex]\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}[/tex]

We have to find the rate of change in volume.

Volume of cube =

[tex]l^3[/tex]

Rate of change of volume =

[tex]\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}[/tex]

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

The rate of change of the volume of the cube at the given instant is; dV/dt = 7.5 m³/h

Surface area is increasing at a rate of 15 m²/h

Thus, dS/dt = 15 m²/h

Now, at a certain instant the surface area of the cube is 24 m².

Formula for cube surface area is; S = 6l²

Thus;

6l² = 24

l² = 24/6

l² = 4

l = √4

l = 2 m

From S = 6l², differentiating both sides with respect to t gives;

dS/dt = 12l(dl/dt)

We know that dS/dt = 15

Thus;

12l(dl/dt) = 15

dl/dt = 15/(12l)

putting 2 for l gives;

dl/dt = 15/(12 * 2)

dl/dt = 0.625 m/h

Now formula for volume of a cube is; V = l³

Differentiating both sides with respect to t gives;

dV/dt = 3l²(dl/dt)

plugging in then relevant values gives;

dV/dt = 3 × 2² × (0.625)

dV/dt = 7.5 m³/h

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