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Thread: Equation

  1. #1
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    Equation

    The temp. in (x, y) is 4*x^2+24*y^2-2*x.
    From all the ellipse points 4*x^2+12*y^2 = 1 find the hotest points and coldest points.
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  2. #2
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    Re: Equation

    I am just rewriting for my own clarification.
    $T(x,y) = 4x^2+24y^2-2x$
    $4x^2+12y^2=1$ This tells us that $-\dfrac{1}{2} \le x \le \dfrac{1}{2}$ (I'm not going to worry about the range for y for now. We can revisit that later).

    So, we want to simplify the function a bit. Let's solve for $y^2$ to get a function of one variable.

    $4x^2+12y^2 = 1$
    $12y^2 = 1-4x^2$
    $y^2 = \dfrac{1-4x^2}{12}$

    Plugging into the formula for $T$, we get:

    $T(x) = 4x^2+24\left(\dfrac{1-4x^2}{12}\right)-2x$
    $T(x) = 4x^2+2-8x^2-2x$
    $T(x) = 2-2x-4x^2 = -4\left(x^2+\dfrac{1}{2}x- \dfrac{1}{2}\right)$
    Now, we can complete the square.
    $T(x) = -4\left(x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)-3$
    $T(x) = -4\left(x+\dfrac{1}{2}\right)^2-3$

    This is a parabola. It reaches its maximum when $x = -\dfrac{1}{2}$. There are at most two values for $y$ for this value of $x$.
    It reaches its minimum when $x$ is as far from $-\dfrac{1}{2}$ as possible. The only value where that can occur is $\dfrac{1}{2}$. There are at most two values for $y$ for this value of $x$.

    Can you find the two or four points, one or two where the temperature is hottest and one or two where it is coldest?
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  3. #3
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    Re: Equation

    Can the problem will be design in something else instead of temperature ?
    Can one replace the temperature in other thing?
    My English o.k. so I write the reply twice.
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