Maximum percentage error in the estimate area of the rectangle

**dangerous_dave** · April 23rd 2008, 06:19 PM

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?

**TheEmptySet** · April 23rd 2008, 06:39 PM

Originally Posted by dangerous_dave

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?

error $=\frac{\Delta A}{A} \approx \frac{dA}{A}$

$A(l,w)=lw$

so

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

so we get

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

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

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