Question 1170709
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The surface area of a cube is  S(a) = 4a^2,  where "a" is the edge size.


We have edge size depending on time  a = a(t);  therefore, the rate of the surface area change is the derivative


    S'(t)  =  4*2*a(t)*a'(t)  =  8*a(t)*a'(t)  {{{cm*(cm/s)}}}.     (1)


The value a'(t) is given : it is  a'(t) = 2 cm/s.


The value of "a" is a= 5, when the volume is 125 cm^3.


Therefore,  the rate of the surface area change is, according the formula (1), 

    S'(t) = 8*5*2 = 80 {{{cm^2/s}}}.     <U>ANSWER</U>
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Solved, &nbsp;answered and explained.   &nbsp;&nbsp;And completed.