Question

If the original 24m edge length x of cube decreases at the rate of 5m/min when x=3m at what rate dose the cube's surface area changes and also volume change​

Answer 1

Answer:

Step-by-step explanation:

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