Question

If the original 24 m edge length x of a cube decreases at the rate 5m/min, when x=3 at what rate does the cube's surface area change

Answer 1

Let s be the surface area and t= time in minutes
s=6x²
d/dt(s)=d/dt(6x²)
ds/dt=12x(dx/dt)
given that x=3, then
ds/dt=12(3)(-5)
ds/dt=-180m²/min

b] Let the volume be v
v=x³
d/dt(v)=d/dt(x³)
dv/dt=3x²(dx/dt)
when x=3, then
dv/dt=3(3²)(-5)
dv/dt=-135m³/min