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Posted by 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 - , 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 - , Sunday, January 2, 2011 at 12:06pm

    thank you so much!!!!

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