Question 1186790
<pre>

These are related-rates problems.

(a)
The surface area of a cube of side a is:
S = {{{ 6*a^2 }}}  

The rate of change of surface area, with respect to t, is:
{{{dS/dt}}} = {{{dS/da}}}{{{da/dt}}}

{{{dS/da}}} = 12a {{{cm}}}
{{{da/dt}}} = 5cm/s   (given)

So we can write:
{{{dS/dt}}} = {{{12a * 5 }}} = <b>{{{60a}}} {{{cm^2/s}}}   (*)</b>

So at a=3cm, the surface area is changing at 
{{{dS/dt}}} = 12(3)cm * 5cm/s = <b>180{{{cm^2/s}}} </b>


(b) plug a=10cm  into (*)