Finding the error in the surface area of a cube. when length = 3, error= ${1\over 4} $

Question

Find the approximate error in the surface area of a cube having an edge of length 3ft if an error of ${1 \over 4}$ in. is made in measuring an edge

I have to do this by using differentials and

This is what I did so far:

Let $x$ be the length

$\Delta x$ be the difference of error

$S$ be the surface area

and 3ft = 36 in.

$$S= 6x^2$$ $$dS = 12x\Delta x \\ = 12 \cdot 36 \cdot {1 \over 4} \\ = 108$$ Is it not $108in.^2$?

It says that it is supposed to be ${9 \over 16} in.^2$

What did I do wrong?

Could anyone show me the steps to the answer? Thanks

Answer 1

I agree with you. An exact calculation is $6(36.25^2-36^2)=108.375\ in^2$ The answer key is close if you are supposed to report the answer in ft$^2$, because $\frac {108}{144}=\frac 34$, but that is still not $\frac 9{16}$

Answer 2

We note that $SA = 6s^{2}.$ However, we represent the error as $\Delta s$ and write $\Delta SA = 6(s + \Delta s)^{2} - 6s^{2}.$ We can expand and simplify this to get $\Delta SA = 12s\Delta s + 6(\Delta s)^{2}.$ Evaluating for $s = 36$ and $\Delta s = \frac{1}{4},$ we get $\Delta SA = 108 + \frac{3}{8}.$