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    I am doing some quantitative analysis on marginal costs and revenue, where I have to find the figure where marginal cost and marginal revenue are equal. I have
    revenue = 720x - 4x^2
    cost = 96-17x^2 + 560x

    Ive got
    marginal revenue = 720 - 8x
    marginal cost = 560 - 34x

    so when marginal cost = marginal revenue

    720 - 8x = 560 - 34x
    26x = -160
    x = -6.15....

    is it even possible to get a negative and what does this mean? at this point is it still maximum profit? which means you have to sell -6 units to make the maximum profit??

    If someone could help today please
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    Somebody who doesn't like milk has negged me. :rolleyes:
    This crappy hand pointing downwards should be removed!
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    (Original post by gff)
    Milk Delivery Problem
    Here's a solution (I hope):

    Spoiler:
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    From the problem statement, we have

    |z^3-z|+|z^5-z^3|+|z^7-z^5|+\cdots + |z^{2013}-z^{2011}| = \sqrt{2012} \ \ \ \ \ (1)
    |z^2-1|+|z^4-z^2|+|z^6-z^4|+\cdots + |z^{2012}-z^{2010}| = \sqrt{2012} \ \ \ \ \ (2)

    From (1),

    |z|(|z^2-1|+|z^4-z^2|+|z^6-z^4|+\cdots + |z^{2012}-z^{2010}|) = \sqrt{2012}

    so from (2),

    |z|\sqrt{2012} = \sqrt{2012} \implies |z| = 1

    Then, from (2),

    |z^2-1|+|z|^2|z^2-1|+|z|^4|z^2-1|+\cdots + |z|^{2010}|z^2-1| = \sqrt{2012}
    1006|z^2-1| = \sqrt{2012}
    |z^2-1| = \frac{\sqrt{2}}{\sqrt{1006}} = \frac{1}{\sqrt{503}}

    Let z = a+bi, where a,b \in \mathbb{R}. Then |z| = 1 implies a^2+b^2 = 1 and |z^2-1| = \frac{1}{\sqrt{503}} implies |(a+bi)^2-1| = \frac{1}{\sqrt{503}}, so |(a^2-b^2-1)+2abi| = \frac{1}{\sqrt{503}} and (a^2-b^2-1)^2+4a^2b^2 = \frac{1}{503}, so a^4+b^4+2a^2b^2-2(a^2-b^2)+1 = (a^2+b^2)^2-2(a^2-b^2)+1 = 1-2(a^2-b^2)+1 = \frac{1}{503}, so a^2-b^2 = \frac{1005}{1006}.

    We have z^2=(a+bi)^2 = (a^2-b^2)+2abi, so \Re (z^2) = a^2-b^2 = \frac{1005}{1006}.

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    (Original post by nohomo)
    Here's a solution (I hope):
    It is a solution.
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    (Original post by nohomo)
    Here's a solution (I hope):

    Spoiler:
    Show


    From the problem statement, we have

    |z^3-z|+|z^5-z^3|+|z^7-z^5|+\cdots + |z^{2013}-z^{2011}| = \sqrt{2012} \ \ \ \ \ (1)
    |z^2-1|+|z^4-z^2|+|z^6-z^4|+\cdots + |z^{2012}-z^{2010}| = \sqrt{2012} \ \ \ \ \ (2)

    From (1),

    |z|(|z^2-1|+|z^4-z^2|+|z^6-z^4|+\cdots + |z^{2012}-z^{2010}|) = \sqrt{2012}

    so from (2),

    |z|\sqrt{2012} = \sqrt{2012} \implies |z| = 1

    Then, from (2),

    |z^2-1|+|z|^2|z^2-1|+|z|^4|z^2-1|+\cdots + |z|^{2010}|z^2-1| = \sqrt{2012}
    1006|z^2-1| = \sqrt{2012}
    |z^2-1| = \frac{\sqrt{2}}{\sqrt{1006}} = \frac{1}{\sqrt{503}}
    [/latex].

    Slightly neater from this point to...

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    Note that |z^2-1| = \sqrt{(z^2-1)(\bar{z}^2-1)} = \sqrt{(z\bar{z})^2 - (z^2 + \bar{z}^2) + 1} = \sqrt{2-(z^2+\bar{z}^2)} so 2 - (z^2 + \bar{z}^2) = \dfrac{1}{503} \implies \dfrac{1005}{1006} = \dfrac{1}{2}(z^2+\bar{z}^2) = \mathrm{Re}(z^2)
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    A fair die is thrown four times and the number of times it falls with a 6 on the top, Y, is noted. Write down all the possible values of y.


    Why is 0 the first answer?
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    (Original post by Farhan.Hanif93)
    Slightly neater from this point to...

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    Note that |z^2-1| = \sqrt{(z^2-1)(\bar{z}^2-1)} = \sqrt{(z\bar{z})^2 - (z^2 + \bar{z}^2) + 1} = \sqrt{2-(z^2+\bar{z}^2)} so 2 - (z^2 + \bar{z}^2) = \dfrac{1}{503} \implies \dfrac{1005}{1006} = \dfrac{1}{2}(z^2+\bar{z}^2) = \mathrm{Re}(z^2)
    That's a good improvement to the solution.

    Does anyone have any more problems for people to solve?
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    (Original post by user)
    A fair die is thrown four times and the number of times it falls with a 6 on the top, Y, is noted. Write down all the possible values of y.


    Why is 0 the first answer?
    What result do you get if you roll 5 4 times?
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    (Original post by nohomo)
    Does anyone have any more problems for people to solve?
    I have tons of fun problems to solve, but, unfortunately, not enough time. Help yourself.
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    I know this is old but I just read the solutions by nohomo (and then the bit farhan posted); thought they were both a bit convoluted (algebra-wise) nearing the end (for me anyway)

    Where nohomo says let z = a+bi

    It's a lot easier to let z^2 = a+bi

    and we have |z^2-1| = \frac{1}{\sqrt{503}}
    or (a-1)^2 + b^2 = 1/503

    and since a^2 + b^2 = 1 then 2a-1 = 1 - 1/503 and then finally a = 1 - 1/1006

    Sorry for the lack of latex; couldn't really be bothered
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    Is this just random questions?

    5x0 = 0
    Re-arrange.
    5 = 0/0

    seems legit
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    (Original post by HypErTwisT)
    Is this just random questions?

    5x0 = 0
    Re-arrange.
    5 = 0/0

    seems legit
    0/0 is an indeterminate. And the answer proves that 0/0 is infinity as it provides no solution, so there can be an arbitrary constant equal to this equation, as in this example:5.
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    Posted for Usycool:

    When perpendicular lines meet, the product of their gradient is -1, right?

    However, say if we have the lines y = 3 and x = 5.

    The gradient of the y=3 in undefined, and the gradient of the x = 5 is 1.

    So is this not an exception for the rule that perpendicular lines's gradients have a product of -1? Or is their another explanation for this?
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    (Original post by GreenLantern1)
    Posted for Usycool:

    When perpendicular lines meet, the product of their gradient is -1, right?

    However, say if we have the lines y = 3 and x = 5.

    The gradient of the y=3 in undefined, and the gradient of the x = 5 is 1.

    So is this not an exception for the rule that perpendicular lines's gradients have a product of -1? Or is their another explanation for this?
    The gradient of the line x=5 is not 1.
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    (Original post by Slumpy)
    The gradient of the line x=5 is not 1.
    I only posted this for my friend not myself (so it has just been copied from what he has written to me). Can you please tell us what all the gradients are and if the prodyct of the gradients doesn't equal -1, then why!!
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    (Original post by GreenLantern1)
    The gradient of the y=3 in undefined, and the gradient of the x = 5 is 1.
    :lolwut:
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    (Original post by olipal)
    :lolwut:
    I didn't agree with that euither - so I just decided to copy and paste what he said to me.


    This isn't my writing but my friends; please just help him out as I am also intriguured by the answer!!!!!!!!!!
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    (Original post by GreenLantern1)
    I didn't agree with that euither - so I just decided to copy and paste what he said to me.


    This isn't my writing but my friends; please just help him out as I am also intriguured by the answer!!!!!!!!!!
    Sorry, it was a typo, I meant the gradient of x = 5 is undefined and the y=3 has gradient 0. My mistake :facepalm:
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    I wonder if I interrupted you guys.

    I've asked this once in the Physics forum but I personally think that this is something about maths as well. I don't know if this is interesting but I guess it will be very challenging:

    Pendulum we meet always have a rigid rod, what if it has a deformable string?

    My approach is like this: I will set a Lagrangian equations like what people did in solving complex pendulum. The deformable rod will be treated as a (very) complicated pendulum with numerous bob with equal mass on it.

    As string can't deform a lot (There is always curvature when you fold a rod), so my solution to the limit of deformation will be modifying the equation so that the bending of that particular point of the rod will have a very high damping, (like k \theta, k will be a constant depending on material and \theta is the bending angle to the normal of that particular point of the rod.)

    So when I can set up an equation, I can gradually increase the points to a very large number and the motion of a pendulum will be simulated.

    Anyway to add the work done against friction into Lagrangian mechanics? And any easier method to do this simulation?

    -Agos
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    (Original post by GreenLantern1)
    I only posted this for my friend not myself (so it has just been copied from what he has written to me).
    That's a bad excuse.

    (Original post by usycool1)
    Sorry, it was a typo, I meant the gradient of x = 5 is undefined and the y=3 has gradient 0. My mistake :facepalm:
    Cool typo.


    Take two lines, y_1 and y_2, and draw them in a similar way as in the picture below.

    Click image for larger version. 

Name:	perpendicular-lines.png 
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    Notice that \tan\theta is the gradient of y_1, and -\tan\varphi is the gradient of y_2. So, the two lines can be expressed in terms of these angles.

    y_1 = \tan(\theta)x, \ \ y_2 = -\tan(\varphi)x

    These will be perpendicular if \displaystyle \theta + \varphi = \frac{\pi}{2}, so you'd use \displaystyle \varphi = \frac{\pi}{2} - \theta.

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    Substituting this, you obtain \displaystyle y_2 = -\tan(\varphi)x = -\tan(\pi/2 - \theta)x = -\cot(\theta)x by co-function identities.

    Therefore, the product of their gradients is -\tan\theta\cot\theta = -1 for a rotation of any angle \theta. Translations also do not change anything.


    In other words, any two perpendicular lines are rotations and translations of the x and y axes, and the product of gradients of those two is -1.

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Updated: October 29, 2014
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