Let's solve the problem **step by step**: --- ### **Given:** - The edge of the cube, \( x \), is **decreasing** at a rate of **5 cm/sec**. So, \( \frac{dx}{dt} = -5 \) cm/s. - Volume of cube: \( V = x^3 \) - Surface area of cube: \( A = 6x^2 \) --- ## **(a) At what rate is the volume of the cube changing when \( x = 2 \) cm?** We want \( \frac{dV}{dt} \) when \( x = 2 \). ### **Step 1: Differentiate volume with respect to time** \[ V = x^3 \implies \frac{dV}{dt} = 3x^2 \frac{dx}{dt} \] ### **Step 2: Substitute the values** - \( x = 2 \) - \( \frac{dx}{dt} = -5 \) \[ \frac{dV}{dt} = 3(2^2)(-5) = 3(4)(-5) = 12(-5) = -60 \] ### **Final Answer:** \[ \boxed{-60\ \text{cm}^3/\text{sec}} \] --- ## **(b) At what rate is the surface area of the cube changing when \( x = 2 \) cm?** We want \( \frac{dA}{dt} \) when \( x = 2 \). ### **Step 1: Differentiate surface area with respect to time** \[ A = 6x^2 \implies \frac{dA}{dt} = 12x \frac{dx}{dt} \] ### **Step 2: Substitute the values** - \( x = 2 \) - \( \frac{dx}{dt} = -5 \) \[ \frac{dA}{dt} = 12(2)(-5) = 24(-5) = -120 \] ### **Final Answer:** \[ \boxed{-120\ \text{cm}^2/\text{sec}} \] --- ## **Summary Table:** | Part | Rate | Value | |------|------|-------| | (a) | \( \frac{dV}{dt} \) | \( -60\ \text{cm}^3/\text{sec} \) | | (b) | \( \frac{dA}{dt} \) | \( -120\ \text{cm}^2/\text{sec} \) | Let me know if you need the answers in a different format!

Certainly! Here are the final answers for both parts, separately: ### **Part (a):** **Rate of change of volume when \( x = 2 \) cm:** \[ \boxed{-60\ \text{cm}^3/\text{sec} \] ### **Part (b):** **Rate of change of surface area when \( x = 2 \) cm:** \[ \boxed{-120\ \text{cm}^2/\text{sec} \]