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27 June, 09:40

What is the maximum area of a right triangle whose legs have a combined length of 10cm?

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  1. 27 June, 09:45
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    1/2ab is the area

    a+b=10

    minus a both sides

    b=10-a

    sub for b in other equation

    1/2a (10-a) = max vol

    5a-1/2a²=max vol

    take deritivive

    5-a=max vol dy/dx

    so it is 0 at a=5

    it changes from positive to negative slope at a=5

    a=5 is the max vol

    then

    b=10-a

    b=10-5

    b=5

    the area is 1/2*5*5=25/2=12.5 square cm
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