Can You Express z in Terms of x and y to Minimize Volume in the First Octant?

[circa415]

Minimizing Volume

If I have a linear plane that cuts through the first octant so that there is a z, y, and x intercept so that you have a triangular face in the octant, is there any way I can put z in terms of x and y? Here's what I'm trying to do

I'm trying to find some linear plane that cuts off the smallest volume in the first octant. The plane must pass through a specific point (let's say it's (2,3,4). I figured since V = Bh and h will stay the same (distance from origin to the point), you will want to minimize the area of the triangular face. So I set each vertice of the triangle as (x,0,0), (0,y,0), and (0,0,z) and take half the cross product of the two vectors <x,0,-z> and <0,y,-z>. But I only know how to find the minimum for only two variables. Is there anyway I could put z in terms of x and y?

I know there is the eq. ax+by+cz+d=0 but I don't see how it would relate to what I'm trying to do right now (finding the mimimum area of the triangular face) and if there is a connection, I don't see it. I think I just use it later on to find the equation for the plane, but right now I'm trying to find the x,y, and z intercepts of the plane..

Or am I just approaching this problem in the wrong way?

Thanks for the help

 

[member 553083]

! Yes, you can use the equation ax + by + cz + d = 0 to find the z-intercept. Substitute x and y with 0, so you get: 0 + 0 + cz + d = 0. Solving for c, you have that c = -d/z. So, your equation for the linear plane is: ax + by -d/z + d = 0. You can use this equation to find the other two intercepts as well.

 

[wais]

!

It is definitely possible to put z in terms of x and y in order to minimize the volume in the first octant. To do this, we can use the equation of the plane, ax+by+cz+d=0, as you mentioned.

First, let's find the equation of the plane that passes through the point (2,3,4). We can do this by plugging in the coordinates of the point into the equation and solving for d. This will give us the equation ax+by+cz+d=0 with d= -2a-3b-4c.

Next, we can use the fact that the plane passes through the x, y, and z intercepts to find the values of a, b, and c. The x intercept occurs when y=0 and z=0, so plugging these values into the equation gives us ax+d=0. Similarly, the y intercept occurs when x=0 and z=0, so plugging these values into the equation gives us by+d=0. Finally, the z intercept occurs when x=0 and y=0, so plugging these values into the equation gives us cz+d=0.

We now have three equations with three unknowns (a, b, and c) and we can solve for them using any method (such as substitution or elimination). Once we have the values of a, b, and c, we can plug them back into the equation ax+by+cz+d=0 to get the equation of the plane in terms of x, y, and z.

This equation will allow us to put z in terms of x and y, and we can use this to minimize the volume in the first octant. We can take the partial derivatives of the volume with respect to x and y, set them equal to 0, and solve for x and y. Then, we can plug these values back into the equation of the plane to get the value of z that minimizes the volume.

In summary, you are on the right track by using the equation of the plane to find the minimum volume. You just need to solve for the values of a, b, and c and use the equation to put z in terms of x and y. Good luck!