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mirajshah
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Homework Statement
What is the maximum possible volume of a rectangular box inscribed in a hemisphere of radius R? Assume that one face of the box lies in the planar base of the hemisphere.
NOTE: For this problem, we're not allowed to use Lagrange multipliers, since we technically haven't learned them yet. I know this is a pain, but please help me out!
Homework Equations
[tex] Volume\, of\, hemisphere\, = \frac{2}{3}R^{3}
\\ Volume\, of\, rectangular\, box=xyz [/tex]
(assuming width x, length y, height z)
The Attempt at a Solution
This is what I've written on my sheet so far:
Assume x,yz is the intersection point of the top vertex of the box with the curved surface of the hemisphere in quadrant I.
[tex] \Rightarrow\text{Dimensions of box}=2x\times2y\times z
\\ \Rightarrow\text{Volume of box}=4xyz [/tex]
If hemisphere has radius R,
[tex] R=\sqrt{x^{2}+y^{2}+z^{2}}
\\ \Rightarrow R^{2}=x^{2}+y^{2}+z^{2} [/tex]
where R is constant for the purposes of maximization.
Therefore, maximize [itex] f\left(x,y,z\right)=4xyz [/itex] with constraint [itex]R^{2}=x^{2}+y^{2}+z^{2}[/itex]
[tex] f_{x}=4yz
\\ f_{y}=4xz
\\ f_{z}=4xy
\\ f_{x}=f_{y}=f_{z}=0
\\ \Rightarrow x=y=z=0 [/tex]
Since this constitutes a non-existant box, the interior critical point is not the global maximum. We must find critical points on the boundary.
Here is where I'm stuck
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