If the surface area of a cube is increasing at a
rate of 3.6 cm2/sec, retaining its shape; then
the rate of change of its volume (in cm3/sec),
when the length of a side of the cube is
10 cm, is :
Solutions · 1
Official solution
For cube of side 'a'
A = 6a2 and V = a3
Given \[{{dA} \over {dt}} = 3.6\]
\[ \Rightarrow \] 12adtda= 3.6{{dV} \over {dt}} = 3{a^2}.{{da} \over {dt}} = 3{a^2}\left( {{{3.6} \over {12a}}} \right) (
at a = 10
\]{{dV} \over {dt}} = 9