Find the area of the largest rectangle that can be inscribed in the ellipse $$fracx^2a^2 + fracy^2b^2 = 1.$$

I gained as far as coming up through the equation for the location to be $A=4xy$ however then once trying to discover the derivative I do not think I"m doing it appropriate.




You are watching: What is the area of the largest rectangle that can be inscribed in the ellipse 4x^2+9y^2=36

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Suppose that the top righthand also edge of the rectangle is at the point $langle x,y angle$. Then you understand that the location of the rectangle is, as you say, $4xy$, and you know that $$fracx^2a^2 + fracy^2b^2 = 1;. ag1$$

Thinking of the area as a function of $x$, we have actually $$fracdAdx=4xfracdydx+4y;.$$ Differentiating $(1)$ through respect to $x$, we have

$$frac2xa^2+frac2yb^2fracdydx=0;,$$ so $$fracdydx=-fracb^2xa^2y;,$$ and $$fracdAdx=4y-frac4b^2x^2a^2y;.$$

Setting this to $0$ and simplifying, we have $y^2=dfracb^2x^2a^2$. From $(1)$ we recognize that $$y^2=b^2-fracb^2x^2a^2;.$$ Therefore, $y^2=b^2-y^2$, $2y^2=b^2$, and also $dfracy^2b^2=dfrac12$. Clearly on, then, $dfracx^2a^2=dfrac12$ too, and also the location is maximized when

$$x=fracasqrt2=fracasqrt22quad extandquad y=fracbsqrt2=fracbsqrt22;.$$


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answered Nov 18 "12 at 22:34
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Brian M. ScottBrian M. Scott
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The vertices of any rectangle inscribed in an ellipse is given by $$(pm a cos( heta), pm b sin( heta))$$ The location of the rectangle is given by $$A( heta) = 4ab cos( heta) sin( heta) = 2ab sin(2 heta)$$ Hence, the maximum is as soon as $sin(2 heta) = 1$. Hence, the maximum location is when $2 heta = dfracpi2$ i.e. $ heta = dfracpi4$. The maximum area is $$A = 2ab$$


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answered Nov 18 "12 at 22:22
user17762user17762
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Gabby Any point on the ellipse if provided by $(a cos( heta), b sin( heta))$. Any quadrilateral inscribed in an ellipse will certainly have works with $(a cos( heta_1), b sin( heta_1))$, $(a cos( heta_2), b sin( heta_2))$, $(a cos( heta_3), b sin( heta_3))$ and $(a cos( heta_4), b sin( heta_4))$. The truth that it is a rectangle enpressures that $ heta_2 = pi - heta_1$, $ heta_4 = - heta_1$ and also $ heta_3 = pi + heta_1$ $endgroup$
–user17762
Nov 18 "12 at 22:33


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$$1=frac x ^ 2 a ^ 2 + frac y ^ 2 b ^ 2 ge frac2 xy ab $$

once and just when $$ x / a = y / b ,$$ the max is got

i.e. :max of $xy =ab/2$, so $4xy=2ab$.


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edited Mar 6 "16 at 10:08
user249332
answered Jan 22 "13 at 6:50
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chenbaichenbai
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let L and H be the length and also breadth of the required rectangle respectively

$$frac(L/2)^2a^2+frac(H/2)^2b^2=1$$

$$frac(L)^24a^2+frac(H)^24b^2=1$$

$$H=fracbasqrt4a^2-L^2$$

Area=L*H

$$A= L*fracbasqrt4a^2-L^2$$

$$fracdAdL =fracbasqrt4a^2-L^2-fracL^2basqrt4a^2-L^2=0$$

$$fracb*(4a^2-2L^2)asqrt4a^2-L^2$$

$$=> 4a^2-2L^2=0$$

$$=2a^2=L^2$$$$L=asqrt2$$$$H=bsqrt2$$$$Area=L*H=2ab$$

$$fracd^2AdL^2=fracsqrt4a^2-L^2*(-4L)-frac4a^2-2L^22sqrt4a^2-L^24a^2-L^2$$

Putting L=$asqrt2$$$fracd^2AdL^2=frac-asqrt2(4asqrt2)-frac02sqrt4a^2-2a^24a^2-2a^2$$

$$=frac-8a^22a^2$$

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