Find the approximate change in the surface area of a cube of side $x$ meters caused by decreasing the side by 1%.

The correct answer is $-0.12 x^2 m^2$

Surface area of a cube = $6x^2$

⇒ $\dfrac{dS}{dx}$ = $12x$ and $\triangle{x} = -1$% of $x$

∴ $\triangle{S} \approx (12x) \triangle{x}$

= $(12x) \left( -\dfrac{x}{100} \right)$

= $-0.12x^2m^2$


∴ $\triangle{S} \approx -0.12 x^2 m^2$