the volume of a cube is increasing at a constant rate of 824 cubic centimeters per minute. At the instant when the volume of the cube is 683683 cubic centimeters, what is the rate of change of the surface area of the cube? Round your answer to three decimal places (if necessary).

DA/dt = 75.27 cm²

Step-by-step explanation:

Cube Volume = V (c) = 683 cm³

DV (c) / dt = 824 cm³

V (c, x) = x³

Then

DV (c, x) / dt = 3x² Dx/dt

(DV (c, x) / dt) / 3x² = Dx/dt (1)

Now as V (c, x) = x³ when V (c, x) = 683 cm³ x = ∛683

x = 8.806 (from excel)

And by subtitution of this value in equation (1)

Dx/dt = (DV (c, x) / dt) / 3x² ⇒ Dx/dt = 824 / 3*x²

Dx/dt = 824 / 3*77.55

Dx/dt = 824/232,64

Dx/dt = 3,542

Then we got Dx/dt where x is cube edge. The area of the face is x² then

the rate of change of the suface area is

DA/dt = (Dx/dt) ²*6 (6 faces of a cube)

DA/dt = (3.542) ²*6

DA/dt = 75.27 cm²