\"Show that the rectangular box of maximum volume inscriptions in a sphere of radius $r$ is a cube\"
I addressed this using Lagrange. However I think this have the right to be fixed using partial derivatives. However, I\"m not completely sure about this. So ns would choose to know if this is possible.
You are watching: Find the maximum volume of a rectangular box that is inscribed in a sphere of radius r.
Sphere is given by $x^2 + y^2 + z^2 = r^2$.
For the rectangle-shaped box with facility at the origin,
$V = 8 xyz = 8xy \\sqrtr^2 - x^2 - y^2$
$\\frac\\partial V\\partial x = 8y \\sqrtr^2 - x^2 - y^2 - \\frac8x^2y\\sqrtr^2 - x^2 - y^2$
To find an essential points, we equate it to zero.
As $\\sqrtr^2 - x^2 - y^2 \\ne 0$ in a sphere, we obtain $8y (r^2 - x^2 - y^2) - 8x^2y = 0$
Again as $y \\ne 0$, $2x^2 + y^2 = r^2$.
Similarly indigenous the various other partial derivative (wrt $y$), we get $2y^2 + x^2 = r^2$
Solving them for $x, y$ and also then plugging right into equation of sphere to find $z$, us will notification they are all equal and also of size $\\frac2r\\sqrt3$.
reply Nov 18 \"20 in ~ 21:54
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