Formula Used: Volume of cube = a3 and Surface area of a cube = 6a2
where a is the side of the cube
Let x be the length of an edge of the cube, V be the volume and S be the surface area at any time t. Then,
V = x3 and S = 6x2
Now, putting the value of V from V = x3, we get,
S = 6x2Differentiating with respect to t, we get,
Thus, the surface area is increasing at the rate of 3.6 cm2/s.
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