The length and width of a rectangle are measured as 30cm and 24 respectively. with an error measurement of at most 0.1cm in each. What is the maximum percentage error in the estimate area of the rectangle?

How do I work this out?

- Apr 23rd 2008, 06:19 PMdangerous_daveMaximum percentage error in the estimate area of the rectangle
The length and width of a rectangle are measured as 30cm and 24 respectively. with an error measurement of at most 0.1cm in each. What is the maximum percentage error in the estimate area of the rectangle?

How do I work this out? - Apr 23rd 2008, 06:39 PMTheEmptySet
error $\displaystyle =\frac{\Delta A}{A} \approx \frac{dA}{A}$

$\displaystyle A(l,w)=lw$

so

$\displaystyle dA=A_l \Delta l + A_w \Delta w$

so we get

$\displaystyle dA=30cm(.1cm)+24(.1cm)cm=5.4cm^2$

error $\displaystyle \approx \frac{5.4cm^2}{720cm^2} = 0.0075= 0.75\%$