Posted by bekah on Sunday, January 2, 2011 at 11:32am.
can you just show me how to set this up? or a formula that could help? im confused.
The length of a rectangle is increasing at 3 ft/min and the width is decreasing at 2 ft/min. When the length is 50 ft and the width is 20 ft, at what rate is the rectangle decreasing?

calc  Reiny, Sunday, January 2, 2011 at 12:00pm
Let the width of the rectangle be x ft
let the length of the rectangle be y ft
Area = xy
d(Area)/dt = x dy/dt + y dx/dt
given: dx/dt = 2 and dy/dt = 3
so when x = 20 and y = 50
d(Area)/dt = 20(3) + 50(2)
= 40
At that moment, the area is decreasing at a rate of 40 ft^2/min 
calc  bekah, Sunday, January 2, 2011 at 12:06pm
thank you so much!!!!