Explanation:
It is given the SURFACE AREA is increasing at 600cm^2 /hr.
(When the edge of the cube is 10 cm.....the surface area is 600 cm^2 )
We need to find how fast the side length is increasing ....then use this to calculate the dV/dt rate
Surface area = L x L x 6
SA = 6 L^2 <===== solve for 'L'
L = sqrt ( 1/6 SA) = (1/6 * SA) ^ 1/2 now differentiate with respect to 't'
dL/dt = 1/12 SA^( -1/2) dSA/dt <==== d SA / dt is given as 600 cm^2 / hr
= 1/12 (600)^( -1/2) * 600 = 2.041 cm/hr <==side length rate of change
volume = L x L x L = L^3 <====== differentiate with respect to 't'
dv/dt = 3 L^2 dL / dt <==== sub in the value we calc'd for dL/dt
3 (10)2 (2.041 ) = 612.3 cm 3 /hr
Let me know if this is incorrect and I will reevaluate !