Question

a rectangle is formed whose length is twice the width. it is enlarged to a similar rectangle as the width changes at a rate of 4 inches per minute. when the width is 10 inches, how fast is the area of the rectangle changing

Answer 1

`(dA)/(dt)=160\ in/min`

Step-by-step explanation:

Given that,

`l=2b`

`(db)/(dt)=4\ in/min`

To find,

Rate of change of area,
`(dA)/(dt)`

Let l is the length and b is the breadth of the rectangle. The area of the rectangle is given by :

`A=l* b`

Differentiating above equation wrt t

`(dA)/(dt)=l(db)/(dt)+b(dl)/(dt)`

`(dA)/(dt)=2b(db)/(dt)+b(dl)/(dt)`

When b = 10 in

`(dl)/(dt)=2(db)/(dt)`

`(dA)/(dt)=2* 10* 4+10* 2(db)/(dt)`

`(dA)/(dt)=2* 10* 4+10* 2* 4`

`(dA)/(dt)=160\ in/min`

So, the area of rectangle is changing at the rate of 160 inches per minute.