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)
let (x,y,z) be the coordinates of the corner of the box that is inscribed in an ellipsoid. Thus "x,y,z \\geq0" and equation of ellipsoid is
"\\frac{x^2}{9}+\\frac{y^2}{16}+\\frac{z^2}{36}=1"
"\\frac{z^2}{36}=1-\\frac{x^2}{9}-\\frac{y^2}{16}"
so "z=6\\sqrt{1-\\frac{x^2}{9}-\\frac{y^2}{16}}"
The volume of the box is represented as
"V=(2x)(2y)(2z)=48xy\\sqrt{1-\\frac{x^2}{9}-\\frac{y^2}{16}}"
For "x\\geq0,y\\geq0, and \\ (\\frac{x^2}{a^2})+(\\frac{y^2}{b^2})\\leq1"
For analysis it is easier to deal with "V^2\\ than\\ V" as it becomes more complex to find the partial derivative in terms of x and y
"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\\ (\\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
"\\frac{\\delta V^2}{\\delta x}=2304[2xy^2-\\frac{4x^3y^2}{9}-\\frac{2x^2y^4}{16})"
"=4608xy^2(1-\\frac{2x^2}{9}-\\frac{y^2}{16})........(1)"
similarly
"\\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
"(1-\\frac{2x^2}{9}-\\frac{y^2}{16})=0\\\\\\frac{2x^2}{9}+\\frac{y^2}{16}=1........(3)"
"(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
"\\frac{2x^2}{9}+\\frac{y^2}{16}=1=\\frac{x^2}{9}+\\frac{2y^2}{16}"
which
"\\frac{x^2}{9}=\\frac{1}{3}\\implies x=\\frac{9}{\\sqrt{3}}"
"\\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=48xy\\sqrt{1-\\frac{x^2}{9}-\\frac{y^2}{16}}"
"=48(\\frac{9}{\\sqrt{3}})(\\frac{16}{\\sqrt{3}})\\sqrt{1-\\frac{1}{3}-\\frac{1}{3}}"
"=1330.215"
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