A optimization problem:
Get the maximum volume of a tetrahedron its 4 vertices on the surface of cube whose edge length is 1 .
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From the geometrical intutition ,we can get :
Selecting $B,A_{1},B_{1},C_{1},$ be the vertices ,then the tetrahedron has maximum volume,and the maximum value is $\frac{1}{3}.$
- But how can I get a strict proof from pure analysis method?（e.g. the method of Lagrange multipliers）
Best Answer
Without loss of generality, take one of the vertices to be the origin. Taking $v_1,v_2,v_3$ to be the coordinates of the remaining vertices, we note that the volume of the tetrahedron is given by $$ V = \frac 16| \det[v_1, v_2, v_3]| $$ This is a quantity on $9$ constrained variables, which you could presumably solve using Lagrange multipliers.