SOLUTION: A cuboid has a square base of side x cm and its height is (2x + 1)cm. Find the approximate increase in its volume when x increases from 10 cm to 10.05 cm. Hence deduce an approxima

Question 1192304: A cuboid has a square base of side x cm and its height is (2x + 1)cm. Find the approximate increase in its volume when x increases from 10 cm to 10.05 cm. Hence deduce an approximate value for the volume when x = 10.05 cm.
Answer by CPhill(2189) (Show Source): You can put this solution on YOUR website!
**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**.

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