Question

The length and width of a rectangle are measured as 55 cm and 49 cm, respectively, with an error in measurement of at most 0.1 cm in each. Use differentials to estimate the maximum error in the calculated area of the rectangle.

Answer 1

The maximum error in the calculated area of the rectangle is
`10.4 \:cm^2`

Explanation:

The area of a rectangle with length
`L` and width
`W` is
`A= L\cdot W` so the differential of A is

`dA=(\partial A)/(\partial L) \Delta L+(\partial A)/(\partial W) \Delta W`

`(\partial A)/(\partial L) = W\\(\partial A)/(\partial W)=L` so

`dA=W\Delta L+L \Delta W`

We know that each error is at most 0.1 cm, we have
`|\Delta L|\leq 0.1`,
`|\Delta W|\leq 0.1`. To find the maximum error in the calculated area of the rectangle we take
`\Delta L = 0.1`,
`\Delta W = 0.1` and
`L=55`,
`W=49`. This gives

`dA=49\cdot 0.1+55 \cdot 0.1`

`dA=10.4`

Thus the maximum error in the calculated area of the rectangle is
`10.4 \:cm^2`