Posted by **Fall** on Wednesday, January 30, 2013 at 4:37pm.

At a certain instant, each edge of a cube is 5 inches long and the volume is increasing at the rate of 2 cubic inches per minute. How fast is the surface area of the cube increasing?

- Calculus -
**Steve**, Wednesday, January 30, 2013 at 5:26pm
a = 6s^2

da/dt = 12s ds/dt

da/dt = 12*5*2 = 120 in^2/min

- Calculus -
**Reiny**, Wednesday, January 30, 2013 at 5:50pm
V = s^3

dV/dt = 3v^2 ds/dt

when s = 5, dV/dt = 2

2 = 3(25) ds/dt

ds/dt = 2/75

A = 6s^2

dA/dt = 12s ds/st

= 12(5) (2/75) = 8/5 inches^2 / min

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