A rectangular box has a square base with an edge length of x cm and a height of h cm. The volume of the box is given by V = x2h cm3. Find the rate at which the volume of the box is changing when the edge length of the base is 4 cm, the edge length of the base is increasing at a rate of 2 cm/min, the height of the box is 15 cm, and the height is decreasing at a rate of 3 cm/min.

  192 cm^3/min

Step-by-step explanation:

Differentiating the volume expression, we get ...

  dV/dt = 2xh(dx/dt) +x^2(dh/dt)

We are given that ...

  x = 4 cm, dx/dt = 2 cm/min, h = 15 cm, dh/dt = -3 cm/min

Putting these values into the formula for volume rate of change, we get ...

  dV/dt = 2(4 cm)(15 cm)(2 cm/min) +(4 cm)^2(-3 cm/min)

  = 240 cm^3/min -48 cm^3/min

  dV/dt = 192 cm^3/min

step-by-step explanation:

first, eliminate the parenthesis:

3

x

+

2

−

4

x

second, group like terms:

3

x

−

4

x

+

2

third, combine like terms:

(

3

−

4

)

x

+

2

→

−

1

x

+

2

→

−

x

+

2