Answer to Question #274756 in Calculus for Hafsa Tahsin

Question #274756

What is the volume of the largest rectangular parallelepiped which can be inscribed in the ellipsoid x2/9+y2/16+z2/36=1? Show that the answer you get gives you the largest volume.(DO NOT USE LAGRANGE MULTIPLIERS)





1
Expert's answer
2021-12-03T13:19:45-0500

let (x,y,z) be the coordinates of the corner of the box that is inscribed in an ellipsoid. Thus x,y,z0x,y,z \geq0 and equation of ellipsoid is

x29+y216+z236=1\frac{x^2}{9}+\frac{y^2}{16}+\frac{z^2}{36}=1


z236=1x29y216\frac{z^2}{36}=1-\frac{x^2}{9}-\frac{y^2}{16}

so z=61x29y216z=6\sqrt{1-\frac{x^2}{9}-\frac{y^2}{16}}


The volume of the box is represented as

V=(2x)(2y)(2z)=48xy1x29y216V=(2x)(2y)(2z)=48xy\sqrt{1-\frac{x^2}{9}-\frac{y^2}{16}}

For x0,y0,and (x2a2)+(y2b2)1x\geq0,y\geq0, and \ (\frac{x^2}{a^2})+(\frac{y^2}{b^2})\leq1


For analysis it is easier to deal with V2 than VV^2\ than\ V as it becomes more complex to find the partial derivative in terms of x and y

V2=2304[x2y2x4y29x2y416)V^2=2304[x^2y^2-\frac{x^4y^2}{9}-\frac{x^2y^4}{16})


Since V=0 if x=0 or y=0 or (x2a2)+(y2b2)=1,V=0 \ if \ x=0 \ or \ y=0 \ or\ (\frac{x^2}{a^2})+(\frac{y^2}{b^2})=1,

the maximum value of V2 and hence of V will occur at critical point of V2 to find the critical point we we have to find the partial derivative in terms of x and y


δV2δx=2304[2xy24x3y292x2y416)\frac{\delta V^2}{\delta x}=2304[2xy^2-\frac{4x^3y^2}{9}-\frac{2x^2y^4}{16})


=4608xy2(12x29y216)........(1)=4608xy^2(1-\frac{2x^2}{9}-\frac{y^2}{16})........(1)


similarly


δV2δy=4608xy2(1x292y216)........(2)\frac{\delta V^2}{\delta y}=4608xy^2(1-\frac{x^2}{9}-\frac{2y^2}{16})........(2)


Now we will equate the above partial derivatives to zero so as to find critical points


(12x29y216)=02x29+y216=1........(3)(1-\frac{2x^2}{9}-\frac{y^2}{16})=0\\\frac{2x^2}{9}+\frac{y^2}{16}=1........(3)


(1x292y216)=0x29+2y216=1........(4)(1-\frac{x^2}{9}-\frac{2y^2}{16})=0\\\frac{x^2}{9}+\frac{2y^2}{16}=1........(4)


Now by equating (3) and (4) we get


2x29+y216=1=x29+2y216\frac{2x^2}{9}+\frac{y^2}{16}=1=\frac{x^2}{9}+\frac{2y^2}{16}


which

x29=13    x=93\frac{x^2}{9}=\frac{1}{3}\implies x=\frac{9}{\sqrt{3}}


y216=13    y=163\frac{y^2}{16}=\frac{1}{3}\implies y=\frac{16}{\sqrt{3}}


The volume of the largest box that can be inscribed inside the ellipsoid is


V=48xy1x29y216V=48xy\sqrt{1-\frac{x^2}{9}-\frac{y^2}{16}}


=48(93)(163)11313=48(\frac{9}{\sqrt{3}})(\frac{16}{\sqrt{3}})\sqrt{1-\frac{1}{3}-\frac{1}{3}}


=1330.215=1330.215



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