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Math Help - Maximise a rectangle inside a triangle

  1. #1
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    Maximise a rectangle inside a triangle

    An isosceles triangle of width and height 4m has a rectangle of height h and width w inside it such that the top corners are always flush to the sides of the triangle. What is the maximum area for the rectangle.

    I know the area of the triangle is 8 and have a function for the area of the rectangle in terms of w and h

    Gives me A(w,h) = 8 - f(w,h)

    The only help in need is another relationship between w and h so I can have the area in terms of one variable either w or h. Then I can make A'=0 , solve that and I will be finished.
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  2. #2
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    Quote Originally Posted by Bushy View Post
    An isosceles triangle of width and height 4m has a rectangle of height h and width w inside it such that the top corners are always flush to the sides of the triangle. What is the maximum area for the rectangle.

    I know the area of the triangle is 8 and have a function for the area of the rectangle in terms of w and h

    Gives me A(w,h) = 8 - f(w,h)

    The only help in need is another relationship between w and h so I can have the area in terms of one variable either w or h. Then I can make A'=0 , solve that and I will be finished.
    Take the bottom-left right-angled triangle or the bottom-right one.

    It's base is \frac{4-w}{2} and it's height is h

    tan\theta=\frac{4}{2}=2=\frac{h}{\left(\frac{4-w}{2}\right)}=\frac{2h}{4-w}
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  3. #3
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    This is using the ratios of the sides of 2 similar triangles?

    Therefore \frac{4}{2} = \frac{h}{2-\frac{1}{2}w}\implies 4-w=h
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  4. #4
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    Yes,

    which is the same as tan(corner\ angle)
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