Question 1192304
**1. Find the Volume Function**

* The volume (V) of the cuboid is given by:
   V(x) = x * x * (2x + 1) 
   V(x) = 2x³ + x²

**2. Find the Initial Volume**

* When x = 10 cm:
   V(10) = 2 * (10)³ + (10)² 
   V(10) = 2000 + 100 
   V(10) = 2100 cm³

**3. Find the Rate of Change of Volume (dV/dx)**

* Differentiate the volume function with respect to x:
   dV/dx = d/dx (2x³ + x²)
   dV/dx = 6x² + 2x

**4. Approximate Increase in Volume**

* Use the differential approximation:
   ΔV ≈ (dV/dx) * Δx 
   where ΔV is the approximate change in volume and Δx is the change in x.

* Δx = 10.05 cm - 10 cm = 0.05 cm

* ΔV ≈ (6 * (10)² + 2 * 10) * 0.05 
   ΔV ≈ (600 + 20) * 0.05 
   ΔV ≈ 620 * 0.05 
   ΔV ≈ 31 cm³

**5. Approximate Volume at x = 10.05 cm**

* Approximate Volume = Initial Volume + Approximate Increase in Volume
* Approximate Volume = 2100 cm³ + 31 cm³
* Approximate Volume = 2131 cm³

**Therefore:**

* The approximate increase in volume when x increases from 10 cm to 10.05 cm is **31 cm³**.
* The approximate volume of the cuboid when x = 10.05 cm is **2131 cm³**.