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  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

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

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answered Nov 18 "12 at 22:34 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 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\$\$

\$\$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\$\$

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

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