(x/a)+(y/b)+(z/c)=1

Question

Find the maximum volume for a rectangular solid (cuboid) with vertices (0,0,0),(x,0,0,),(x,y,0),(0,y,0),(0,0,z),(x,0,z),(0,y,z)and (x,y,z) with this last vertex on the plane                    (x/a)+(y/b)+(z/c)=1

Answer 1

Hi Salman,

 

One method for solving these types of problems is called Lagrange Multiplier method.

 

Let f (x, y, z) = xyz (the volume of the cuboid), and g (x, y, z) = x/a + y/b + z/c, so g (x, y, z) = 1 for the vertex whose coordinates aren't fixed. Set ∇f = k∇g, where k is an unknown scalar. You get 3 equations in addition to the fourth equation g (x, y, z) = 1.

 

yz = k/a

xz = k/b

xy = k/c

 

Divide the second equation by the first to get

x/y = a/b

 

Divide the third equation by the first to get

x/z = a/c

 

Cross multiply in both equations to get

x/a = y/b and x/a = z/c.

 

This shows x/a = y/b = z/c, and we know x/a + y/b + z/c = 1, so each summand is 1/3. Therefore x = a/3, y = b/3, z = c/3.

 

This shows the volume achieves a max or min at (a/3, b/3, c/3). Since the volume is everywhere continuous, it must achieve its global max somewhere on the portion of the plane x/a + y/b + z/c = 1 that's in the positive orthant. This can't happen on the boundary b/c along the boundary one or more of x, y, z are 0. This means the max occurs at the point above.