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  • The length x, of a rectangle is decreasing at the rate of 5 cm/minute and the width y, is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rate of change of the area of the rectangle.  

The length x, of a rectangle is decreasing at the rate of 5 cm/minute and the width y, is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rate of change of the area of the rectangle.

 

 

 

 
 
 
 
 

Answers (1)

    \frac{dx}{dt}= -5 cm/min ---(i)
&  \frac{dy}{dt}= 4 cm/min ---(ii)
Now area of the rectangle , A = xy
\Rightarrow \frac{dA}{dt}= x\frac{dy}{dt}+y\frac{dx}{dt}
\frac{dA}{dt}= x \left ( 4 \right )+y\left ( -5 \right )\; \; \left [ using (i) and (ii) \right ]
\therefore \frac{dA}{dt}= 4x-5y
when x = 8cm  and y = 6cm
\therefore \left [ \frac{dA}{dt} \right ]_{at\, x= 8,y= 6}           = 4\left ( 8 \right )-5\left ( 6 \right )
                                           = 32-30= 2cm^{2}/min
Hence , the rate of change of the area of the rectangle is 2cm2 / min.

Posted by

Ravindra Pindel

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