Results 1 to 4 of 4

Math Help - Maximizing/Minimizing the area of a rectangle

  1. #1
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381

    Maximizing/Minimizing the area of a rectangle

    HOw can you maximize/minimize the area of a rectangle inscribed in a triangle?
    See the figure below w/ given measurements of the triangle.
    10 , 8 and 6
    Help please...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    HOw can you maximize/minimize the area of a rectangle inscribed in a triangle?
    See the figure below w/ given measurements of the triangle.
    10 , 8 and 6
    Help please...
    Choose a single parameter which describes the size of the rectangle (see the attachment), and calculate the area of the rectangle in terms of that single parameter.

    So here:

    A=(3/4) x(8-x)

    which you are to find the maximum of (for x in [0,8]).

    RonL
    Attached Thumbnails Attached Thumbnails Maximizing/Minimizing the area of a rectangle-gash.jpg  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by CaptainBlack View Post
    So here:

    A=(3/4) x(8-x)

    which you are to find the maximum of (for x in [0,8]).

    RonL
    We do this by differentiating A with respect to x, setting the derivative to zero, and solving for x. This is the x which gives the maximum (or other stationary point type, but in this case it is a maximum) area.

    A=(3/4) (8x-x^2)

    dA/dx=(3/4)(8-2x)

    so dA/dx=0, gives 2x=8, or x=4.

    Plugging x=4 back into the formula for the area gives the maximum area of a rectangle inscribed in the given triangle is:

    A=(3/4) 4(8-4)=12.

    and the actual dimensions of the rectangle can be found from the relations between u, v x and the dimensions of the triangle shown in the diagram in my previous post

    RonL
    Last edited by CaptainBlack; February 11th 2007 at 03:59 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    HOw can you maximize/minimize the area of a rectangle inscribed in a triangle?
    See the figure below w/ given measurements of the triangle.
    10 , 8 and 6
    Help please...
    Here is one way.
    We will use your figure as posted.

    Let x = width of the inscribed rectangle
    And y = length.

    On the figure, on the horizontal leg of the 6-8-10 right triangle, or on the 6-unit long leg. Let's call that horizontal leg, line segment AB, where
    point A is the corner of the big triangle at the right angle
    point C is the corner of the big triangle, at the other end of the AB.

    Call the 3rd corner of the big triangle, the one at the top, as point C.
    Call the corner of the inscribed rectangle along CA as point D.
    Call the corner of the inscribed rectangle along AB as point E.
    Call the lower corner of the inscribed rectangle along BC as point F.

    Right triangles DAE and CAB are similar, so, are proportional.
    DE/CB = AE/AB
    y/10 = AE/6
    AE = 6y/10 = 3y/5 ------------**

    Right triangles EFB and CAB are similar, so, are proportional.
    EB/CB = EF/CA
    EB/10 = x/8
    EB = 10x/8 = 5x/4 ----------**

    Now,
    AE +EB = AB
    3y/5 +5x/4 = 6
    Clear the fractions, multiply both sides by 5*4,
    12y +25x = 120
    12y = 120 -25x
    y = (120 -25x)/12
    y = 10 -25x/12 ---------------(i)

    Area of inscribed rectangle, A = x*y
    A = x(10 -25x/12)
    A = 10x -(25x^2)/12
    Differentiate both sides with respect to x,
    dA/dx = 10 -(1/12)(50x)
    dA/dx = 10 -25x/6
    Set dA/dx to zero,
    0 = 10 -25x/6
    x = 10 / (25/6)
    x = 60/25 = 12/5 = 2.4 units long -------------for max A.
    So,
    y = 10 -25x/12 = 10 -25*2.4/12 = 10 -5 = 5 units long ----for max A.

    Therefore, to maximize the area of the inscribed rectangle, positioned as shown on the figure, make the width = 2.4 units long, and the length = 5 units long. -----------------answer.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 12th 2010, 11:32 PM
  2. Maximizing and Minimizing Area
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 12th 2009, 02:14 PM
  3. Replies: 1
    Last Post: September 28th 2009, 05:00 PM
  4. Replies: 2
    Last Post: March 21st 2008, 10:43 AM
  5. Maximizing and minimizing
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: October 16th 2007, 02:30 PM

Search Tags


/mathhelpforum @mathhelpforum