To solve this problem, let’s break down each part and calculate the required values using differentials.

Given:

• Edge length of the cube, $x = 15$ cm.
• Possible error in measurement, $dx = 0.1$ cm.

We need to find:

1. The maximum possible error, relative error, and percentage error in computing:
   • (a) the volume of the cube.
   • (b) the surface area of the cube.

(a) Volume of the Cube

The volume $V$ of a cube with edge length $x$ is given by: $V = x^3$

1. Differential of Volume

To estimate the maximum possible error in the volume, we use the differential: $dV = 3x^2 \cdot dx$

Plugging in the values: $dV = 3 \times (15)^2 \times 0.1 = 3 \times 225 \times 0.1 = 67.5 \, \text{cm}^3$

So, the maximum possible error in volume is $67.5 \, \text{cm}^3$.

2. Relative Error in Volume

The relative error is given by: $\text{Relative Error} = \frac{dV}{V}$

Calculating $V$ with $x = 15$: $V = 15^3 = 3375 \, \text{cm}^3$

Then, $\text{Relative Error} = \frac{67.5}{3375} \approx 0.02$

So, the relative error in volume (rounded to four decimal places) is $0.0200$.

3. Percentage Error in Volume

The percentage error is simply the relative error multiplied by 100: [ \text{Percentage Error}