Question

The original 24 m edge length x of a cube decreases at the rate of 2 ​m/min. a. When x=4 ​m, at what rate does the​ cube's surface area​ change? b. When x=4 ​m, at what rate does the​ cube's volume​ change?

Option 1: Surface area rate
Option 2: Volume rate
Option 3: Both surface area and volume rates
Option 4: No rate change

Answer 1

a. The rate of change of the cube's surface area when x = 4 m is -96 m^2/min. b. The rate of change of the cube's volume when x = 4 m is -96 m^3/min.

Step-by-step explanation:

a. To find the rate of change of the cube's surface area, we can use the formula for the surface area of a cube which is given by S = 6x^2. Differentiating with respect to time gives dS/dt = 12x(dx/dt).

Plugging in the values x = 4 and dx/dt = -2, we get dS/dt = 12(4)(-2) = -96 m^2/min.

Therefore, the rate of change of the cube's surface area when x = 4 m is -96 m^2/min.

b. To find the rate of change of the cube's volume, we can use the formula for the volume of a cube which is given by V = x^3. Differentiating with respect to time gives dV/dt = 3x^2(dx/dt).

Plugging in the values x = 4 and dx/dt = -2, we get dV/dt = 3(4)^2(-2) = -96 m^3/min.

Therefore, the rate of change of the cube's volume when x = 4 m is -96 m^3/min.