# Thread: Maximum percentage error in the estimate area of the rectangle

1. ## Maximum 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?

2. 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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# calculate the percentage error in the area of a rectangle

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