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Maximize the area of a triangle

  1. Jul 18, 2015 #1

    Consider the semicircle with radius 1, the diameter is AB. Let C be a point on the semicircle and D the projection of C onto AB. Maximize the area of the triangle BDC.

    What I'm thinking
    y=sqrt(r^2-x^2) From the formula of a circle x^2+y^2=r^2
    A=1/2(x+1)y The area of a triangle
    A'=(1-x-2x^2)/2sqrt(1-x^2) Substitution from the formula of a circle and take the derivative
    A'=0 if x=1/2 Set the derivative = to 0 to maximize it
    y=sqrt(3)/2 Solve for y
    A=1/2((1/2)+1)sqrt(3)/2= 3sqrt(3)/8 Plug in all the pieces into the original equation of the area of a triangle.

    A=Area A'=Derivative of area
    x+1 = base of the triangle
    y=The height

    These are still new to me let me know how far off I am. I think I have the equation wrong.

    (sorry posted this in wrong area first)
    Last edited: Jul 18, 2015
  2. jcsd
  3. Jul 18, 2015 #2


    Staff: Mentor

    The above is very difficult to read or make sense of. What are x and y?
    Is r2 supposed to mean r2? If so, you should at least write r^2 to indicate that 2 is an exponent.
    Homework problems should be posted in the Homework & Coursework section, using the homework template.
  4. Jul 18, 2015 #3

    Ray Vickson

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    Science Advisor
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    The working may be OK (depending on what you mean by the various symbols), but is spoiled by a careless presentation.

    First: what is 'x', and what is its range? (Don't make the reader guess---just tell him/her what you mean).
    Second: when you write r2 and x2, do you mean ##r^2## and ##x^2##? If so, write them as r^2 and x^2 in plain text (which is what you would also need to do for numerous software packages if you were using a computer to calculate things).
    Third: use parentheses around leading fractions like your '1/2'; in other words, write (1/2)(x+1) y ---- or better, use (1/2)*(x+1)*y in plain text. Parentheses help to avoid confusion between 1/2ab (which can be interpreted as ##\frac{1}{2ab}## and (1/2) ab (which, unambiguously, means ##\frac{1}{2} ab##).
    Finally----and maybe most importantly--- break up your sequence of equations into separate lines, or at least, make it clear what goes with what. The way you wrote it looks like
    [tex] y = \sqrt{r2-x2}A=1/2(x+1)y A' = (1x - 2x2)/2 \sqrt{1-x2} A'=0 [/tex]
    instead of the intended (I hope) meaning
    [tex] y = \sqrt{r^2 - x^2} = \sqrt{1-x^2} \;(\text{since} \: r = 1).[/tex]
    The area is
    A = (1/2)(x+1) A, \;\text{so that the derivative is} \; A' = (x - 2 x^2)/(2 \sqrt{1-x^2}).[/tex]
    The maximal area occurs when ##A' = 0##, or ##x = 1/2##, giving maximal area ##A = 3 \sqrt{3}/8##.

    You don't need to use LaTeX (as I have done), but you do need to put in occasional clarifying words/sentences so that others can follow what you are trying to do. Even the simple strategy of putting different things on different lines will help, so you could write
    A'=0 if x=1/2, y=sqrt(3)/2, A=1/2((1/2)+1)sqrt(3)/2= 3sqrt(3)/8.

    That could get you higher marks, too.
  5. Jul 18, 2015 #4
    I fixed it not sure what the template is but I gave the question and my attempt, What else should I conclude?

    ^ thank you you broke down what I was trying to say. I'm not aware of latex I will use it in the future I"m on a phone so not sure I can use it though. The x was just the extra distance to get the base of the triangle on top of the radius. That is the part I think I'm messing up with also.
    Last edited: Jul 18, 2015
  6. Jul 18, 2015 #5

    Ray Vickson

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    BTW: You don't need quite so many parentheses, so you don't need to say sqrt(r^(2)-x^(2)); just plain sqrt(r^2 - x^2) will do, because the parsing rules for mathematical expressions will take care of the "precedence". You still need to develop the habit of defining your terms: saying what are x, y and A for example. (In the case of x, it is particularly important, because two different people doing this problem might well use two different definitions of x!)
  7. Jul 18, 2015 #6
    Thanks for the feedback I think I tried to define everything a little better in the original post.
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