Without using Lagrange multipliers, find the maximum volume of a rectangular box inscribed in tetrahedron bounded by the coordinate planes and the plane 2x/5 + y + z = 1I am self-teaching vector calculus but I got stuck on this question. Thanks for any help! :)

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Savanah Hernandez · Accepted Answer

Geometrically, you can view your rectangular box as the parallelepiped spanned by three vectors lying on the positive coordinate axes, which can therefore be written as   x      i  ,   y      j  , and   z      k   for some numbers   x,   y, and   z. The only vertex that doesn&#039;t lie on a coordinate axis or plane is the vertex furthest from the origin, which therefore has coordinates   (  x  ,  y  ,  z  ); being inscribed by your tetrahedron, then, is the same as requiring that this vertex lie on the given plane, which is the same as requiring that             2      5        x  +  y  +  z  =  1. As a result, you&#039;re maximising the volume of the box, which is the function  V  (  x  ,  y  ,  z  )  =  x  y  z  ,subject to the constraint            2      5        x  +  y  +  z  =  1At this point, you could throw Lagrange multipliers at the problem, but this is total overkill, since you can just solve the constraint equation for   z, say, rewrite the volume of the box as a function of just   x and   y, and now maximise this function of two variables on the base of the tetrahedron in the   x  y-plane.

Armeninilu · Accepted Answer

we don&#039;t need Lagrange multipliers. To maximize   x  y  z subject to             2      5        x  +  y  +  z  =  1, use the AM-GM inequality:            2      5        x  ⋅  y  ⋅  z  ≤            (                        1          3                    (                        2          5                    x      +      y      +      z      )      )        3    =            1      27          ⟹    x  y  z  ≤            5      54        .

Without using Lagrange multipliers, find the maximum volume of a

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