Calculus
posted by Anna .
If the volume of a cube is increasing at 24 in^3/min and the surface area of the cube is increasing at 12 in^2/min, what is the length of each edge of the cube?
I know that dV/dt=24 and ds/ds=12. I also know that Volume=s^3 and Surface Area=6*s^2, but what do I do from there?

from V = x^3
dV/dt = 3x^2 dx/dt
dx/dt = 24/(3x^2) = 8/x^2
from A = 6x^2
dA/dt = 12x dx/dt
dx/dt = 12/(12x) = 1/x
then 8/x^2 = 1/x
x^2 = 8x
x = 8 
Let the independent variable be
s = length of one side of the cube.
As you mentioned,
V(s) = s³
S(s) = 6s²
dV/dt = dV/ds*ds/dt = 3s² ds/dt = 24 ... (1)
dS/ds = dS/ds*ds/dt = 12s ds/dt = 12 ...(2)
Solve for s and ds/dt by substitution or dividing (1) by (2). 
Let the independent variable be
s = length of one side of the cube.
As you mentioned,
V(s) = s³
S(s) = 6s²
dV/dt = dV/ds*ds/dt = 3s² ds/dt = 24 ... (1)
dS/ds = dS/ds*ds/dt = 12s ds/dt = 12 ...(2)
Solve for s and ds/dt by substitution or dividing (1) by (2).
I get s=8, and ds/dt=1/8. 
8

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