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Math Help - trig problem

  1. #1
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    trig problem

    Homework Problem:

    What is the largest possible area for a parallelogram that has pairs of sides with lengths 5 and 11?

    R(subscript)max=?

    Thanks
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  2. #2
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    Re: trig problem

    Quote Originally Posted by jonathanraxa View Post
    Homework Problem:

    What is the largest possible area for a parallelogram that has pairs of sides with lengths 5 and 11?

    R(subscript)max=?

    Thanks
    The area of the parallelogram is

    \displaystyle \begin{align*} A &= a\,b\sin{\theta} \\ &= 5 \cdot 11 \cdot \sin{\theta} \\ &= 55\sin{\theta} \end{align*}

    Since this is a sine function that has not been translated vertically, its maximum value is equal to its amplitude.
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    Re: trig problem

    Area of parallelogram= base * height Take 11 as the base. Height will be maximum when the parallelogram is a rectangle. So max area=11*5=55
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    Re: trig problem

    Quote Originally Posted by biffboy View Post
    Area of parallelogram= base * height Take 11 as the base. Height will be maximum when the parallelogram is a rectangle. So max area=11*5=55
    Which of course is when \displaystyle \begin{align*} \theta = 90^{\circ} \end{align*} because that maximises the sine function, which I wrote in my post
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