1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculus: Optimization Problem

  1. Apr 3, 2008 #1
    Calculus: Optimization Problems

    1. The problem statement, all variables and given/known data
    Find the area of the largest rectangle that can be inscribed in the ellipse below.
    [​IMG]

    I'm not quite sure where to start.....first of all, how would you even enter this into a calculator to graph? On the TI-83, I only see one variable 'x' that you can enter...

    I was reading a similar problem that said this was a lagrange multiplier problem. I was never taught this method. Is there any other way to do this?

    ________________________________________________________________________________________________________
    2. The upper right-hand corner of a piece of paper, 14 in. by 10 in., as in the figure, is folded over to the bottom edge. How would you fold it so as to minimize the length of the fold? In other words, how would you choose x to minimize y?
    [​IMG]
    Heres what I tried to do:

    First I used the pythagorean theorem. a^2+b^2=y^2
    We're trying to minimize y, so I set up the distance for each side.
    (14-b)=top
    (10-x)=side

    Plug:
    (14-b)^2+(10-x)^2=y^2

    Since we can't solve for two variables, solve for one. Now what would I do? or would both variables be just "x"? like:
    (14-x)=top
    (10-x)=side

    Thanks for the help!
     
    Last edited: Apr 3, 2008
  2. jcsd
  3. Apr 3, 2008 #2

    How abt assuming four points[tex] (a cos\theta , b sin \theta), (a cos\theta ,-b sin \theta),(-a cos\theta , b sin \theta),(-a cos\theta ,-b sin \theta) [/tex]

    to be the vertices of that rectangle....then area =[tex]4absin\theta cos\theta=2absin 2\theta[/tex]

    and then maximizing the area ? So 2ab ?
     
  4. Apr 3, 2008 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Your idea of starting a problem is to enter it into a calculator? Surely you should be able to graph an ellipse on paper and draw a rectangle in it?

    The hard part would be using symmetry to show that the largest rectangle must have its sides parallel to the axes of the ellipse and that the rectangle is symmetric about both axes but you probably are expected to assume that. Assuming that, The coordinates of the vertices of the rectangle are (x0,y0), (-x0,y0), (-x0,-y0), and (x0,-y0) ( have subscripted to distinguish from the "variables" x and y). What is the area of such a rectangle, as a function of x0 and y0? Of course, x0 and y0 must also satisfy
    [tex]\frac{x_0^2}{a^2}+ \frac{y_0^2}{b^2}= 1[/itex]
    Yes, "Lagrange multiplier" would be the simplest way to solve that but you can also use the equation of the ellipse to reduce to one variable.
    (Simple answer, by the way.)
     
  5. Apr 3, 2008 #4
    This seems to be a pretty wrong equation to me. Which triangle are we concerned about here? (14-b) means, length-breadth. This can hardly mean any side of a triangle.

    What you need to note for the 2nd problem is that, for a given 'x', only a particular value of 'y' is possible i.e. 'y' is a function of 'x' and 'x' alone. Try to arrive at 'y(x)'.
     
  6. Apr 3, 2008 #5
    Another approach:
    My mathematics handbook gives the parametric form of an ellipse as:
    x(t) = a cos t;
    y(t) = b sin t;
    The area of the inscribed rectangle is
    area(t) = 4 x(t) y(t);
    Solving the differential equation d/dt area(t)=0 gives the value of t at at the maximum
    area. Using this value gives the area(t)=2 a b as shown in a previous reply.
     
  7. Apr 5, 2011 #6
    Re: Calculus: Optimization Problems


    1.
    Can't see 1st image (inaccessible) :(


    2.
    Hint:
    Hypotenuse in right-angled triangle - minimum - isoscles.






    =>
    Solution:
    x=
    10
    y=
    14.1421356





    Explanation:

    If done practically, the solution presents itself instantaneously.
    http://goo.gl/Cirv3




    Theoretically:
    The triangle formed by the folded paper is a right-angled triangle (diag).
    The minimum hypotenuse in a right-angled triangle is when both the sides are equal (isoscles).
    so y is minimum when 10-x=0 i.e. when x=
    10

    so
    y2 = x2 + x2
    y2= 2 . x2
    y= x . [tex]\sqrt[]{}[/tex]2
    y = 10 (1.41421356)
    y=14.14
     
  8. Oct 9, 2011 #7
    Aruna is (respectfully) wrong.
    The real minimum occurs when x=7.5, which gives y=12.99
    This can be found legitimately (as opposed to aruna's hand-wavy argument) using similar triangles.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Calculus: Optimization Problem
Loading...