Results 1 to 5 of 5

Math Help - Optimizing a rectangle

  1. #1
    Newbie
    Joined
    Feb 2009
    Posts
    5

    Optimizing a rectangle

    A rectangle has a constant area of 200 square meters and its length L is increasing at the rate of 4 meters per second
    1)Find the width W at the instant the width is decreasing at the rate of .5 meters per second.
    2) At what rate is the diagonal D of the rectangle changing at the instant when the width W is 10 meters?

    1)So I know that DL/Dt= 4 and Dw/dt= -.5
    So the equation is A=LW, the derivative being Da/dt= L(dw/dt)+W(dL/dt)
    Would that be 0=L(-.5)+W(4)
    What would I do next there?

    2)How would I start this one off?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Jan 2009
    Posts
    108
    1) excellent start. now replace either w with (200/l) or replace l with (200/w) in your derivative equation, then solve for the one remaining variable.

    2) the length of the diagonal, if you forget the formula, can easily be rediscovered by using a theorem by a fellow named pythagoras. go ahead, draw a diagonal in and find it's length in terms of l and w.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2009
    Posts
    5
    Wow, thanks, I cant believe I forgot that the area was 200. So when I sub that in I get w=5.
    For the diagonal the equation would be L^2 + W^2 = D^2
    2L(dl/dt)+2w(dw/dt)=2D(dd/dt)
    200=WL = 200=10(L)
    L=20

    20^2+10^2=D^2
    D=√500

    A=LW
    0=20dw/dt+10dL/dt
    -20dw/dt=10(4)
    dw/dt=-2
    2(20)(4)+2(10)(-2)=√500(dd/dt)
    so i got dd/dt to round out to about 5.367?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member OnMyWayToBeAMathProffesor's Avatar
    Joined
    Sep 2006
    Posts
    157

    reply

    I too got w=5 for this problem. But i got something else for the rate of the diagonal.

    This is my work:

    w=10 so if the area is 200m^2 then L must equal 20. But the area now is 100m^2 because it is a triangle. so with the formula x^2+y^2=d^2 where x=width=10 and y=height=20 and d=hypotenuse=\sqrt{500}. The derivative of that would then be 2x\frac{dx}{dt}+2y\frac{dy}{dt}=2d{dd}{dt}. Thus when we plug everything in, we get, 2(10)\frac{dx}{dt}+2(20)\frac{4}{1}=2\sqrt{500}\fr  ac{dd}{dt}.

    to find \frac{dx}{dt} we do A=\frac{1}{2}xy which the derivative of is A'=\frac{1}{2}x\frac{dy}{dt}+\frac{1}{2}y\frac{dx}  {dt}. when we plug everything in, we get 0=\frac{1}{2}10\frac{4}{1}+\frac{1}{2}20\frac{dx}{  dt} which equals 0=20+10\frac{dx}{dt} = -20=10\frac{dx}{dt} so \frac{dx}{dt}=-2.

    if we plug -2 back into 2(10)\frac{dx}{dt}+2(20)\frac{4}{1}=2\sqrt{500}\fr  ac{dd}{dt}, \frac{dd}{dt}=2.683.

    did i make any mistakes anywhere? I feel like i did, but the logic makes sense. any reactions?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,695
    Thanks
    450
    Quote Originally Posted by calc_student09 View Post
    A rectangle has a constant area of 200 square meters and its length L is increasing at the rate of 4 meters per second
    1)Find the width W at the instant the width is decreasing at the rate of .5 meters per second.
    2) At what rate is the diagonal D of the rectangle changing at the instant when the width W is 10 meters?

    1)So I know that DL/Dt= 4 and Dw/dt= -.5
    So the equation is A=LW, the derivative being Da/dt= L(dw/dt)+W(dL/dt)
    Would that be 0=L(-.5)+W(4)
    What would I do next there?

    you know LW = 200 ...

    2)How would I start this one off?

    how about D^2 = L^2 + W^2 ?
    .
    Last edited by skeeter; March 18th 2009 at 05:55 PM. Reason: way too slow ...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: June 2nd 2011, 04:56 PM
  2. Replies: 3
    Last Post: March 24th 2010, 06:32 PM
  3. Replies: 1
    Last Post: November 23rd 2009, 06:41 PM
  4. Optimizing
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 1st 2009, 01:50 PM
  5. Optimizing a rectangle
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 18th 2009, 04:30 PM

Search Tags


/mathhelpforum @mathhelpforum