Question

The volume of a cube is increasing at a rate of 10cm^3/min. How fast is the surface area increasing when the length of an edge is 30cm?

Answer 1

\( \cfrac{dV}{dt}=+10\qquad \qquad V=lwh\qquad
\begin{cases}
l=x\\
w=x\\
h=x
\end{cases}\implies V=x\cdot x\cdot x\implies V=x^3\)

in a cube, all sides are equal, thus length, height and width are all "x", thus you'd end up with something like \(x^3\)

Answer 2

let x be the edge of the cube
volume \[V=x^3\]
\[\frac{ dV }{ dt }=3x^2\frac{ dx }{ dt },\]
when x=30 cm
\[10=3*30^2\frac{ dx }{ dt }\]
\[\frac{ dx }{ dt }=?\]
surface area \[S=6x^2\]
\[\frac{ dS }{ dt }=6*2x \frac{ dx }{ dt }\]
when x=30 cm
find \[\frac{ dS }{ dt }\]

Answer 3

yeap, notice above as surjithayer said



the cube is just 6 squares, each of x * x, so the surface area of it will be \(\bf s = 6x^2\)