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Math Help - I think this is hard....

  1. #1
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    I think this is hard....

    In arch  OA\frown of the line in the plane , with equation f(x)=x^3 where O(0,0) and A(a,a^3) for a>0. Find the coordinates of the point M(X0,Y0) in order that \triangle OAM has the biggest area , than find the equation of the tangent and normal line in point M
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  2. #2
    MHF Contributor red_dog's Avatar
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    The area of the triangle is

    S=\frac{1}{2}|\Delta| where \Delta=\begin{vmatrix}x_O & y_O & 1\\x_A & y_A & 1\\x_M & y_M & 1\end{vmatrix}

    \Delta=\begin{vmatrix}0 & 0 & 1\\a & a^3 & 1\\x & x^3 & 1\end{vmatrix}=x^3-a^2x

    Then S(x)=\frac{a}{2}|x^3-a^2x|=\left\{\begin{array}{ll}\frac{a}{2}(a^2x-x^3), & x\in(-\infty,-a]\cup [0,a]\\<br />
\frac{a}{2}(x^3-a^2x), & x\in(-a,0)\cup(a,\infty)\end{array}\right.

    S'(x)=\left\{\begin{array}{ll}\frac{a}{2}(a^2-3x^2), & x\in(-\infty,-a)\cup (0,a)\\<br />
\frac{a}{2}(3x^2-a^2), & x\in(-a,0)\cup(a,\infty)\end{array}\right.

    S'(x)=0\Rightarrow x=\pm\frac{a\sqrt{3}}{3} and both are maximum points.

    Then, the maximum area is S\left(\pm\frac{a\sqrt{3}}{3}\right)
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  3. #3
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    Quote Originally Posted by red_dog View Post

    Then S(x)=\frac{a}{2}|x^3-a^2x|=\left\{\begin{array}{ll}\frac{a}{2}(a^2x-x^3), & x\in(-\infty,-a]\cup [0,a]\\<br />
\frac{a}{2}(x^3-a^2x), & x\in(-a,0)\cup(a,\infty)\end{array}\right.

    S'(x)=\left\{\begin{array}{ll}\frac{a}{2}(a^2-3x^2), & x\in(-\infty,-a)\cup (0,a)\\<br />
\frac{a}{2}(3x^2-a^2), & x\in(-a,0)\cup(a,\infty)\end{array}\right.

    S'(x)=0\Rightarrow x=\pm\frac{a\sqrt{3}}{3} and both are maximum points.

    Then, the maximum area is S\left(\pm\frac{a\sqrt{3}}{3}\right)
    Can you explain me a little bit please, because i can't understand it.
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