Find the dimensions of a rectangular box of maximum value th

Find the dimensions of a rectangular box of maximum value that has three of its faces in the coordinate planes, one vertex at the origin and another vertex in the first octant on the plane 4x + 3xy + z = 12.

Solution

Let the vertex in the first octant be (x;y;z). Then the recatangular box has the side length x,y and z respetively. Thus its volume is
V = x?y?z
Furthermore the poinr has to be on the plane:
4?x +- 3?y + z = 12

Instead of using Lagrange multiplier to combine this constraint with the obejective function V, you can use thiis equation to express one variable as function for the two other. When you apply this result to V you get a new objective of two variabkles, which can be maximized.

E.g. solve for z :
z = 12 - 4?x - 3?y
So
V = x?y?(12 - 4?x - 3?y) = 12?x?y - 4?x