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 to 10.05 cm. Hence deduce an approximate va
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Question 1192389: 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 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 of the Cuboid**
* **Original Volume (V):**
V = Base Area × Height
V = x² × (2x + 1)
When x = 10 cm:
V = 10² × (2 * 10 + 1)
V = 100 × 21
V = 2100 cubic cm
**2. Approximate Increase in Volume**
* **Use Differentials:**
* Volume (V) = x² * (2x + 1) = 2x³ + x²
* dV/dx = 6x² + 2x
* Approximate change in volume (dV) ≈ (dV/dx) * dx
where dx is the change in x (0.05 cm in this case)
* dV ≈ (6 * 10² + 2 * 10) * 0.05
* dV ≈ (600 + 20) * 0.05
* dV ≈ 31 cubic cm
**3. Approximate Volume when x = 10.05 cm**
* **Approximate New Volume:**
* New Volume ≈ Original Volume + Approximate Increase in Volume
* New Volume ≈ 2100 cubic cm + 31 cubic cm
* New Volume ≈ 2131 cubic cm
**Therefore:**
* **Approximate increase in volume:** 31 cubic cm
* **Approximate volume when x = 10.05 cm:** 2131 cubic cm
This method provides an approximation of the volume change using differentials. For a more precise calculation, you could directly calculate the volume at x = 10.05 cm and compare it to the original volume.