The volume of a cube is increasing at the rate of `8 c m^3//s`. How fast is the surface area increasing when the length of an edge is 12 cm?

Explanation

Let edge of cube x, volume be V and surface area be S.
`:. V = x^(3) and S = 6x^(2)`
Given `(dV)/(dt) = 8 cm^(3) //sec`
`V= x^(3) rArr (dV)/(dt) = 3x^(2) (dx)/(dt)`
` rArr 8= 3x^(2) (dx)/(dt) rArr (dx)/(dt) = 8/(3x^(2))` …(1)
Again `S = 6 x^(2)`
` rArr (dS)/(dt) = 12x (dx)/(dt)`
` 12x xx 8/(3x^(2))` From equatin (1)
`=32/x=32/12" "(.:' x= 12 cm)`
`8/3 cm^(2)//sec`