# Percentage error: linear approximation

I have trouble understanding the question below and I do not really know what linear approximation has to do with this:

Determine how accurate should we measure the side of a cube so that the calculated surface area of the cube lies within 3% of its true value, using Linear Approximation.

Let $A(x)= TSA$; $x=side$

$A(x)=6x^2$

And since we only want the change in $A(x)$ within $3\%$,

$\frac{dA}{dx}= \frac{3}{100}\left(12x\right)$

From here what can I do? Or am I taking a wrong procedure?

Suppose the true value of the side is $s$, the linear approximation says that $$A(x) \approx A(s) + \frac{dA(s)}{d x}.(x-s) = A(s) + 12 s (x-s)$$ The relative error on the area's calculation is $$\frac{A(x)-A(s)}{A(s)} \approx \frac{12 s (x - s)}{A(s)} = \frac{2(x-s)}{s}$$ that is to say twice the relative error on the side's measure. In order to have $$\big|\frac{A(x)-A(s)}{A(s)}\big|\le 0.03$$ we need $$\big|\frac{x-s}{s}\big|\le 0.015$$ that is to say $1.5\%$
• Suppose the side is $s = 10$, the area is $A(s) = 600$. Instead of $10$, you measure $x = 10.5$. You compute an area of $A(x) = 661.5$. The (absolute) error is $|661.5 - 600| = 61.5$. The relative error (or error in percentage) is $61.5/600 = 0.1025 = 10.25\%$. – Gribouillis Oct 21 '17 at 10:42