Answer:
[tex]\dfrac{dA}{dt}=160\ in/min[/tex]
Explanation:
Given that,
[tex]l=2b[/tex]
[tex]\dfrac{db}{dt}=4\ in/min[/tex]
To find,
Rate of change of area, [tex]\dfrac{dA}{dt}[/tex]
Let l is the length and b is the breadth of the rectangle. The area of the rectangle is given by :
[tex]A=l\times b[/tex]
Differentiating above equation wrt t
[tex]\dfrac{dA}{dt}=l\dfrac{db}{dt}+b\dfrac{dl}{dt}[/tex]
[tex]\dfrac{dA}{dt}=2b\dfrac{db}{dt}+b\dfrac{dl}{dt}[/tex]
When b = 10 in
[tex]\dfrac{dl}{dt}=2\dfrac{db}{dt}[/tex]
[tex]\dfrac{dA}{dt}=2\times 10\times 4+10\times 2\dfrac{db}{dt}[/tex]
[tex]\dfrac{dA}{dt}=2\times 10\times 4+10\times 2\times 4[/tex]
[tex]\dfrac{dA}{dt}=160\ in/min[/tex]
So, the area of rectangle is changing at the rate of 160 inches per minute.
