Asked by bekah
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?
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?
Answers
Answered by
Reiny
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 <b>decreasing</b> at a rate of 40 ft^2/min
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 <b>decreasing</b> at a rate of 40 ft^2/min
Answered by
bekah
thank you so much!!!!
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.