please, I have got elipse which is definied $4x^2+9y^2=36$ and I should determine the sizes of $a,b$ so that the rectangle placed into elipse has maximum area. Thanks for help.
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The maximum size of a rectangle inside a circle $4x^2+4y^2=36$ is a square $3\times3=9$.
$\therefore$ The maximum size of a rectangle inside an ellipse $4x^2+9y^2=36$ is a rectangle $3\times2=6$.
Kay K.
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Hint:
First note that a rectangle can be inscribed in an ellipse also if its sides are parallel to the symmetry axis.
Now, a point $P$ on the ellipse has coordinates $P=(x_P,y_P)=(x,\pm\frac{2}{3} \sqrt{9-x^2})$. Chosing a point in the first quadrant, with $x_P>0$ and $y_P>0$ the area of the rectangle is $A=4x_Py_P$ so you have to find the maximum of the function:
$$ f(x)=\frac{8}{3} x\sqrt{9-x^2} $$
can you do this?
Emilio Novati
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Yes it is. 3/sqrt(2). But I do not know next steps. – Peter Dec 11 '15 at 13:31
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Right! Substitute this value in $f(x)$ and you have the area. If you want the two sides $a,b$ of the rectangle, note that $a=2x_P$ and $b=2y_P$. – Emilio Novati Dec 11 '15 at 19:29
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Please. I have got the area which is 12, but really struggle to find a and b. Please can you give me last hind. thanks a lot – Peter Dec 11 '15 at 23:54
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I have the problem with the equation. The sizes should be sqrt(2)3 and sqrt(2)2. But do not know how to get there. – Peter Dec 12 '15 at 14:49
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You have. $a=2\cdot \frac{3}{\sqrt{2}}=3\cdot\sqrt{2}$ and $b=2\cdot \frac{2}{3}\sqrt{9-\frac{9}{2}}=\frac{4}{3}\frac{3}{\sqrt{2}}=2\sqrt{2}$. – Emilio Novati Dec 12 '15 at 14:55