Question

The length and width of a rectangle are measured as 30cm and 24cm, respectively, with an error in measured of

Answer 1

Answer:

The maximum error in the calculated area of rectangle is 5.4 cm².

Step-by-step explanation:

Given : The length and width of a rectangle are measured as 30 cm and 24 cm, respectively, with an error in measured of  at most 0.1 cm in each.

To find : Use differentials to estimate the maximum error in the calculated area of rectangle ?

Solution :

The area of the rectangle is $A=l\times w$

The derivative of the area is equal to the partial derivative of area w.r.t. length times the change in length plus the partial derivative of area w.r.t. width times the change in width.

i.e. $dA=\frac{\partial A}{\partial L}\Delta L+\frac{\partial A}{\partial W}\Delta W$

Here, $\frac{\partial A}{\partial L}=30,\ \frac{\partial A}{\partial W}=24 ,\ \Delta L=0.1,\ \Delta W=0.1$

Substitute the values,

$dA=(30)(0.1)+(24)(0.1)$

$dA=3+2.4$

$dA=5.4$

Therefore, the maximum error in the calculated area of rectangle is 5.4 cm².