Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I'm haveing optimization problems

  1. Apr 6, 2004 #1
    This is homework (forgive me) but I don’t want an answer I would just like to know what I am doing wrong.

    Here is the problem:

    Find the area of the largest rectangle that can be inscribed in a right triangle with legs of lengths 3cm and 4cm if tow dies of the rectangle lie along the legs.

    Here is what I did

    L = length of rectangle
    W = width
    Theta = angle 4 leg of the right triangle

    I’m trying to optimize W*L where :
    W < 3
    L < 4

    These are the equations I got:
    Tan(theta) = w/(4-L)
    Tan(theta) = (3-W)/L
    Tan(theta) = 3/4

    W = [itex] 3(4-L)/4 [/itex]


    {3 - 3(4-L)/4}/ L = 3/4

    and that is as far as this brain will take me...
    Last edited: Apr 7, 2004
  2. jcsd
  3. Apr 7, 2004 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Basically, what you have done "wrong" is back-substituting, so that you end up with the trivially correct equation 3/4=3/4 .
    You already have a perfectly good expression for the width, w=3(4-L)/4.
    With this expression, what is the area of the rectangle?
    How can you find the maximum of this area?
  4. Apr 7, 2004 #3
    After you figure out that one, try this one.

    Basically the same problem, but one side of the rectangle is on the hypotenuse of the triangle.
  5. Apr 7, 2004 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Set up a coordinate system with the right angle at, (4,0) thus the line representing the hypotenuse is given by y= 3x/4.

    Let L and H be the sides of your rectangle. One corner must be on the line y(x) = 3x/4
    one side will be L=4-x the other H=y

    The area is A=L*H = (4-x)*y = (4-x)*3x/4

    Compute [tex] \frac {dA} {dx}= 0 [/tex]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook