Answer:
The maximum error in the calculated area of the rectangle is [tex]10.4 \:cm^2[/tex]
Step-by-step explanation:
The area of a rectangle with length [tex]L[/tex] and width [tex]W[/tex] is [tex]A= L\cdot W[/tex] so the differential of A is
[tex]dA=\frac{\partial A}{\partial L} \Delta L+\frac{\partial A}{\partial W} \Delta W[/tex]
[tex]\frac{\partial A}{\partial L} = W\\\frac{\partial A}{\partial W}=L[/tex] so
[tex]dA=W\Delta L+L \Delta W[/tex]
We know that each error is at most 0.1 cm, we have [tex]|\Delta L|\leq 0.1[/tex], [tex]|\Delta W|\leq 0.1[/tex]. To find the maximum error in the calculated area of the rectangle we take [tex]\Delta L = 0.1[/tex], [tex]\Delta W = 0.1[/tex] and [tex]L=55[/tex], [tex]W=49[/tex]. This gives
[tex]dA=49\cdot 0.1+55 \cdot 0.1[/tex]
[tex]dA=10.4[/tex]
Thus the maximum error in the calculated area of the rectangle is [tex]10.4 \:cm^2[/tex]
