Results 1 to 5 of 5

Math Help - Help

  1. #1
    Newbie
    Joined
    Jun 2014
    From
    Virginia
    Posts
    2

    Help

    consider a symmetric cross inscribed in a circle of radius r

    http://img593.imageshack.us/img593/5237/cross.png

    a) Write the area A of the cross as a function of x and find the value of x that maximizes the area.
    b) Write the area A of the cross as a function of θʹ and find the value of θʹ that maximizes the area.
    c) Show that the critical numbers of parts (a) and (b) yield the same maximum area. What is the area?
    Attached Thumbnails Attached Thumbnails Help-untitled.jpg  
    Last edited by saintsguy25; June 29th 2014 at 07:38 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    May 2008
    From
    Melbourne Australia
    Posts
    216
    Thanks
    29

    Re: Help

    Let y be half the thickness of one arm. Then y^2 = r^2 - x^2

    So the thickness of one arm is 2\sqrt{(r^2-x^2)}

    The area of the cross is twice the area of each large rectangle less the double counted square in the middle:

    A=8x\sqrt{(r^2-x^2)}-4y^2

    A=8x\sqrt{(r^2-x^2)}-4r^2+4x^2

    \frac{dA}{dx}=8\sqrt{(r^2-x^2)}+4x(r^2-x^2)^{-\frac 12}(-2x)+8x=0

    \therefore 8(r^2-x^2)-8x^2+8x\sqrt{(r^2-x^2)}=0

    \therefore r^2-2x^2+x\sqrt{(r^2-x^2)}=0

    \therefore (\frac rx)^2-2+\sqrt{(\frac rx)^2 -1}=0

    \therefore \sqrt{(\frac rx)^2 -1}=2-(\frac rx)^2

    \therefore (\frac rx)^2 -1=4-4(\frac rx)^2+(\frac rx)^4

    \therefore (\frac rx)^4-5(\frac rx)^2+5=0

    (\frac rx)^2=\frac{5 \pm \sqrt{25-20}}{2}

    I hope this is right! haven't had much time to check it.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2014
    From
    Virginia
    Posts
    2

    Re: Help

    How would you suggest doing part b and c?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    May 2008
    From
    Melbourne Australia
    Posts
    216
    Thanks
    29

    Re: Help

    One way would be to just substitute x=r cos(\theta) and y=r sin(\theta) into

    A=8x\sqrt{(r^2-x^2)}-4y^2

    then simplify and follow the same procedure as for part a.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,932
    Thanks
    782

    Re: Help

    I disagree with Kiwi_Dave's solution for part (b). The problem gives you \theta, which will not be the same as if you set x = r\cos \theta, y = r\sin \theta. In this instance, \theta is a constant.

    Let's give labels to the intersections of the cross with the circle. Starting at the point (1,0) and going counterclockwise, the intersections of the cross with the circle are A,B,C,D,E,F,G,H. Let's call the center of the circle O. By symmetry, the areas of triangles OBC, ODE, OFG, and OHA are all equal. To calculate the area of triangle OBC, it is the following area:

    \dfrac{r^2}{2}\sin \theta

    Next, to figure out the area we have not counted yet requires we figure out the angle AOB. By symmetry, this angle will be \dfrac{2\pi - 4\theta}{4} = \dfrac{\pi}{2}-\theta. So, the area we are missing in the first quadrant is the area of triangle AOB minus the area of a right triangle. That right triangle has sides of length x - y.

    So, let's calculate the area of triangle AOB:
    \dfrac{r^2}{2}\sin\left(\dfrac{\pi}{2}-\theta\right) = \dfrac{r^2}{2}\cos \theta

    Next, x = r\cos\left(\dfrac{\theta}{2} \right), y = r\sin\left(\dfrac{\theta}{2}\right), so the right triangle we wish to remove has area:

    \dfrac{r^2}{2}\left(\cos\left(\dfrac{\theta}{2} \right) - \sin\left(\dfrac{\theta}{2} \right)\right)^2 = \dfrac{r^2}{2}(1-\sin \theta)

    So, the area we are missing is the difference of these two:
    \dfrac{r^2}{2}\left(\cos \theta + \sin \theta -1 \right)

    So, the total area as a function of \theta is

    A(\theta) = 4r^2\sin \theta + 2r^2\cos \theta - 2r^2

    Then, taking the derivative to find critical numbers:

    A'(\theta) = 4r^2\cos \theta -2r^2\sin \theta

    This has critical numbers when \tan \theta = 2. This gives \sin \theta = \dfrac{2}{\sqrt{5}}, \cos \theta = \dfrac{1}{\sqrt{5}}.

    To show that this is the same critical value you found for x, you just need to convert from \theta to x. You have x = r\cos\left(\dfrac{\theta}{2}\right). The half-angle formula for cosine is \cos\left(\dfrac{\theta}{2}\right) = \sqrt{\dfrac{1+\cos \theta}{2}}. So,

    x = r\sqrt{\dfrac{1+\tfrac{\sqrt{5}}{5}}{2}} = r\sqrt{\dfrac{5+\sqrt{5}}{10}}

    Solving what Kiwi_Dave found a bit more, you get the same answer.
    Last edited by SlipEternal; July 1st 2014 at 07:38 AM.
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum