# Thread: Using La-Grange Multiplier to find volume of constrained box

1. ## Using La-Grange Multiplier to find volume of constrained box

I don't know how I would go about doing this question. I am revising for an exam and have gotten stuck. Please Help meh

2. ## Re: Using La-Grange Multiplier to find volume of constrained box

The first step is to find the x and y coordinates of the corners for a given z. And to do that you need to work out a formula for the cone. Here' what I would do: Looking just at the first octant, a cross section at the x,z plane gives a line from x= 0, z= 1 to x= 2, z= 0. That is given by z= 1- x/2 or x= 1- 2z. Similarly, a cross section at the y,z plane gives a line from y= 0, z= 1 to y= 1, z= 0. That is given by z= 1- y or y= 1- z.

So at a given "z", the cross section is an ellipse with semi-axes 1- 2z and 1- z. The equation of that ellipse is $\frac{x^2}{(1-2z)^2}+ \frac{y^2}{(1- z)^2}= 1$.

Now we can say that we want to maximize the volume, V= xyz, subject to the constraint $A= \frac{x^2}{(1-2z)^2}+ \frac{y^2}{(1- z)^2}= 1$

To solve that using "Lagrange multipliers" set $\nabla V= \lambda\nabla A$.

That, together with the constraint, give four equations for x, y, z, and $\lambda$.

3. ## Re: Using La-Grange Multiplier to find volume of constrained box

at z = 0 , your equation gives a cross section

$x^2 + y^2 = 1$

which is a circle. It should be an ellipse.

The cross section should be

$x^2+4y^2=4(z-1)^2$

and the volume = 4xyz