Lagrange Multipliers, calc max volume of box

In summary, the point P(x,y,z) lies on the ellipsoid 2x^2 + 10y^2 + 5z^2 = 80 in the first octant of space and is a vertex of a rectangular parallelpiped. Using the Method of LaGrange Multipliers, the coordinates of P can be determined so that the box has a maximum volume. The maximum volume can also be calculated using the equation f(x,y,z)=xyz and the constraint g(x,y,z)=2x^2+10y^2+5z^2=80. The final solution is y=sqrt(8/3), x=sqrt(40/3), z=sqrt(16/3).
  • #1
hils0005
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Homework Statement



Point P(x,y,z) lies on the part of the ellipsoid 2x^2 + 10y^2 + 5z^2 = 80 that is in the first octant of space. It is also a vertex of a rectangular parallelpiped each of whose sides are parallel to a coordinate plane. Use Method of LaGrange Multipliers to determine the coordinates of P so that the box has a max volume and calculate the max



Homework Equations


f(x,y,z)=xyz g(x,y,z)=2x^2+10y^2+5z^2=80


The Attempt at a Solution



[tex]\nabla[/tex]f=[tex]\nabla[/tex]g[tex]\lambda[/tex]

1.yz=4x[tex]\lambda[/tex]
2.xz=20y[tex]\lambda[/tex]
3.xy=10z[tex]\lambda[/tex]

I multiplied equation 1 by x, 2 by y and 3 by z

4x^2[tex]\lambda[/tex]=20y^2[tex]\lambda[/tex]=10z^2[tex]\lambda[/tex]

I then put x and z in terms of y and put into constraint
4x^2=20y^2 10z^2=20y^2
x=[tex]\sqrt{}[/tex]5 y z=[tex]\sqrt{}[/tex]2 y

g=2([tex]\sqrt{}[/tex]5y)^2 +10y^2 + 5([tex]\sqrt{}[/tex]2y)^2=80
solving for y=[tex]\sqrt{}[/tex](8/3)

I'm not sure if I'm on the right track or if this is way off, if correct do I just do the same proceedure to find x and z?
 
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  • #2
You are doing fine. You don't need to repeat the procedure for x and z. You already have x=sqrt(5)*y and z=sqrt(2)*y. Once you've got y, you've got everything.
 

1. What are Lagrange Multipliers and how are they used?

Lagrange Multipliers are a mathematical tool used to find the maximum or minimum value of a function subject to constraints. In the context of finding the maximum volume of a box, Lagrange Multipliers are used to optimize the volume while keeping the surface area of the box constant.

2. Why is the maximum volume of a box important to calculate?

The maximum volume of a box is important to calculate in many real-world applications, such as optimizing storage space or designing packaging. It allows for the most efficient use of space while still meeting certain constraints, resulting in cost and resource savings.

3. How do you set up the equation for finding the maximum volume of a box using Lagrange Multipliers?

The equation for finding the maximum volume of a box using Lagrange Multipliers is set up by first defining the function to optimize (in this case, volume) and the constraint function (in this case, surface area). These two functions are then combined into a single equation, with a Lagrange Multiplier added to account for the constraint. This equation is then solved for the values of the variables that result in the maximum volume.

4. Can Lagrange Multipliers be used to find the minimum volume of a box?

Yes, Lagrange Multipliers can be used to find the minimum volume of a box by following the same process as finding the maximum volume. However, in this case, the Lagrange Multiplier will account for the minimum constraint, such as a minimum required volume or a maximum cost for materials.

5. Are there any limitations to using Lagrange Multipliers for optimizing the volume of a box?

While Lagrange Multipliers are a powerful tool for optimization, there are some limitations to their use. They may not work for all types of constraints, and in some cases, the resulting solution may not be the global maximum or minimum. It is important to carefully consider the constraints and the overall problem before using Lagrange Multipliers to find the maximum volume of a box.

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