Vector Calculus - maximization

Question

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 = 1

I am self-teaching vector calculus but I got stuck on this question. Thanks for any help! :)

Answer 1

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 $x \mathbf{i}$, $y \mathbf{j}$, and $z \mathbf{k}$ for some numbers $x$, $y$, and $z$. The only vertex that doesn'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 $\tfrac{2}{5}x + y + z =1$. As a result, you're maximising the volume of the box, which is the function $$ V(x,y,z) = xyz, $$ subject to the constraint $$ \tfrac{2}{5}x + y + z =1 $$ At 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 $xy$-plane.

Answer 2

As @BranimirĆaćić noted, we don't need Lagrange multipliers. To maximize $xyz$ subject to $\tfrac25x+y+z=1$, use the AM-GM inequality:$$\tfrac25x\cdot y\cdot z\le\left(\tfrac13(\tfrac25x+y+z)\right)^3=\tfrac{1}{27}\implies xyz\le\tfrac{5}{54}.$$