Results 1 to 8 of 8

Math Help - Objective Functions

  1. #1
    Junior Member
    Joined
    Sep 2008
    From
    Canada
    Posts
    42

    Objective Functions

    Express the area of an equilateral triangle as a function of its height h.

    Stuck on this quite a bit..I'm terrible at word problems.
    A=h ?
    h=x

    I'm confused...I need to plug "H" into the area..

    A=1/2bh...but how do I get "H"? Or should I solve for the base and then plug it in to get my objective function?
    Last edited by tiar; September 28th 2008 at 02:13 PM. Reason: copied text wrong
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,318
    Thanks
    1232
    Quote Originally Posted by tiar View Post
    Express the area of an equilateral triangle as a function of its height h.

    Stuck on this quite a bit..I'm terrible at word problems.
    A=h ?
    h=x

    I'm confused...I need to plug "H" into the area..

    A=1/2bh...but how do I get "H"? Or should I solve for the base and then plug it in to get my objective function?
    It's hard to solve without drawing a diagram - so as I explain this to you, draw it.

    Say you have an equilateral triangle of side length l. Clearly all the angles are 60^{\circ}.

    If you're expressing the Area in terms of his height, then we need to find an expression for the height. So draw the height, remembering that the height is perpendicular to the base, and will cut one of the angles in half.

    So in other words, an equilateral triangle is the same as two right-angled 30, 60, 90 degree triangles put next to each other.

    Just looking at one of the right angled triangles, we know that the hypotenuse must be of length l and one of the side lengths must be \frac{1}{2}l. We can use pythagoras to find the height.

    a^2 + b^2 = c^2

    \left (\frac{1}{2}l \right)^2 + h^2 = l^2

    \frac{1}{4}l^2 + h^2 = l^2

     h^2 = \frac{3}{4}l^2

     h = \frac{\sqrt{3}}{2}l .


    Rearranging for l we get l = \frac{2 \sqrt{3}}{3} h.


    So now we can say...

    A = \frac{1}{2}bh.

    Since b = l = \frac{2\sqrt{3}}{3}h we get...

    A = \frac{1}{2} \times \frac{2\sqrt{3}}{3}h \times h

    and so...


    A = \frac{\sqrt{3}}{3}h^2 .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2008
    From
    Canada
    Posts
    42
    Quote Originally Posted by Prove It View Post
    It's hard to solve without drawing a diagram - so as I explain this to you, draw it.

    Say you have an equilateral triangle of side length l. Clearly all the angles are 60^{\circ}.

    If you're expressing the Area in terms of his height, then we need to find an expression for the height. So draw the height, remembering that the height is perpendicular to the base, and will cut one of the angles in half.

    So in other words, an equilateral triangle is the same as two right-angled 30, 60, 90 degree triangles put next to each other.

    Just looking at one of the right angled triangles, we know that the hypotenuse must be of length l and one of the side lengths must be \frac{1}{2}l. We can use pythagoras to find the height.

    a^2 + b^2 = c^2

    \left (\frac{1}{2}l \right)^2 + h^2 = l^2

    \frac{1}{4}l^2 + h^2 = l^2

     h^2 = \frac{3}{4}l^2

     h = \frac{\sqrt{3}}{2}l .

    Hi, I am wondering how you went from...
    \frac{1}{4}l^2 + h^2 = l^2
    to...
     h^2 = \frac{3}{4}l^2

    What am I not thinking of...I do not get the \frac{3}{4}...
    Thank you for your response, I greatly appreciate it!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1
    Tiar
    The post by Prove_It is correct.
    Go with it.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Sep 2008
    From
    Canada
    Posts
    42
    Quote Originally Posted by Plato View Post
    Tiar
    The post by Prove_It is correct.
    Go with it.
    I know it is..I am merely wondering how he got from the 1/4 to 3/4. I haven't done math in years...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    h^2 = \frac{4l^2}{4} - \frac{1l^2}{4}

    which is

    h^2 = l^2\bigg(\frac{4}{4} - \frac{1}{4}\bigg)

    h^2 = \frac{3}{4}l^2
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Sep 2008
    From
    Canada
    Posts
    42
    Quote Originally Posted by 11rdc11 View Post
    h^2 = \frac{4l^2}{4} - \frac{1l^2}{4}

    which is

    h^2 = l^2\bigg(\frac{4}{4} - \frac{1}{4}\bigg)

    h^2 = \frac{3}{4}l^2

    Oh! Thank you! For some reason I thought I was multiplying it. Goes to show how I am too tired to think...
    This makes perfect sense now! Thank you!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,318
    Thanks
    1232
    Quote Originally Posted by Plato View Post
    Tiar
    The post by Prove_It is correct.
    Go with it.
    Now now Plato, no need to be like that. Everyone has lapses of memory now and then. It's always the easy stuff too that gets forgotten the most easily.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Multi-objective recursive least squares
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: May 11th 2010, 07:46 PM
  2. [SOLVED] objective quantities
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 20th 2007, 09:09 AM
  3. Objective Function
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: November 17th 2007, 08:58 PM
  4. Objective functions, region of feasible solutions
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: April 22nd 2007, 05:36 PM
  5. max profit and objective functions
    Posted in the Business Math Forum
    Replies: 3
    Last Post: September 28th 2006, 06:23 AM

Search Tags


/mathhelpforum @mathhelpforum