I think this is hard....

**beq!x** · April 25th 2009, 10:55 AM

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 $M$

**red_dog** · April 25th 2009, 12:01 PM

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 $S\left(\pm\frac{a\sqrt{3}}{3}\right)$

**beq!x** · April 26th 2009, 03:23 AM

Originally Posted by red_dog

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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