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Math Help - I have a couple of parabola and quadratic equation related issues.

  1. #1
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    I have a couple of parabola and quadratic equation related issues.

    Hello again MHF!

    Just a couple of questions that I don't know how to work out fully.
    *******************************
    Find the dimensions of a rectangle a with the greatest area whose perimeter is 30 feet.

    I have a plan to work it out--I think.
    First I have to divide the standard perimeter formula by 2 in order to get: l+w=15. Then solve for 'l' in terms of 'w'(l=15-w), plug that into the area formula:A=w(15-w) ,which is equal to,A=15w-w^2.
    The thing is, I have no idea how to find the maximum possible value for the length, width, and area.
    I think it might be that I have to use the area formula I noted above(A=15w-w^2) as a formula for a parabola and find the maximum of it, but I'm not too sure if I'm right.
    I don't even know how to find the maximum of an equation in that form....
    ********************************
    Given x^3-4x^2+2x+1=0:
    (a) how many possible positive roots are there?
    (b) how many possible negative roots are there?
    (c) what are the possible rational roots?
    (d) using synthetic substitution, which of the possible rational roots is actually a root of the equation?
    (e)find the irrational roots of the equation.(hint: use the quadratic formula to solve the depressed equation.)

    I'm really clue less as to solving this one.....I got lost after possible rational roots.
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  2. #2
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    Quote Originally Posted by koumori View Post
    Hello again MHF!

    Just a couple of questions that I don't know how to work out fully.
    *******************************
    Find the dimensions of a rectangle a with the greatest area whose perimeter is 30 feet.

    I have a plan to work it out--I think.
    First I have to divide the standard perimeter formula by 2 in order to get: l+w=15. Then solve for 'l' in terms of 'w'(l=15-w), plug that into the area formula:A=w(15-w) ,which is equal to,A=15w-w^2.
    The thing is, I have no idea how to find the maximum possible value for the length, width, and area.
    I think it might be that I have to use the area formula I noted above(A=15w-w^2) as a formula for a parabola and find the maximum of it, but I'm not too sure if I'm right.
    I don't even know how to find the maximum of an equation in that form....

    If you know how to, differentiate. If you don't, find the vertex of the parabola. The parabola given by your area equation is downward facing, so it's vertex is the maximum value it could take.

    ********************************
    Given x^3-4x^2+2x+1=0:
    (a) how many possible positive roots are there?
    (b) how many possible negative roots are there?
    (c) what are the possible rational roots?
    (d) using synthetic substitution, which of the possible rational roots is actually a root of the equation?
    (e)find the irrational roots of the equation.(hint: use the quadratic formula to solve the depressed equation.)

    I'm really clue less as to solving this one.....I got lost after possible rational roots.

    So I assume you're fine with parts a,b,and c of this question?

    Look at your list of rootsBy inspection, you can easily see that x = 1 is a root. Divide the original polynomial by (x-1). You'll get a quadratic.

    Use the quadratic formula on the quadratic you found in (e) to find the irrational roots

    .
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  3. #3
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    First question:

    <br />
\underset{x}{\fbox{A=xy}}y



    A=xy
    P=2x+2y
    y=\frac{P-2x}{2}

    \therefore

    A=x\frac{30x-2x^2}{2}

    then: if you know how:

    \frac{DA}{Dx}\left [x(\frac{p-2x}{2}) \right ] =

    \frac{1}{2}\frac{DA}{Dx}\left [(px-2x^2) \right ]
    \frac{1}{2}(p-4x)

    \frac{P}{2}-2x=0
    <br />
x=\frac{P}{4}
    <br />
x=\frac{30}{4}

    x=7.5



    If you don't know differentiation:


    \frac{Px-2x^2}{2}=-x^2+15x

    Use \frac{-b}{2a}

    \frac{-15}{-2}=7.5

    this is your x, plug into your equation to get y and those are the values to gain the max area.
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