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OleMash [197]
3 years ago
15

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 c

m in each. Use differentials to estimate the maximum error in the calculated area of the rectangle.
Mathematics
1 answer:
valentina_108 [34]3 years ago
4 0

Answer:

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

Step-by-step explanation:

The area of a rectangle with length L and width W is A= L\cdot W so the differential of <em>A</em> is

dA=\frac{\partial A}{\partial L} \Delta L+\frac{\partial A}{\partial W} \Delta W

\frac{\partial A}{\partial L} = W\\\frac{\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

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