Does a Minimum Exist for x^2+y^2+z^2 Given x^4+y^4+z^4=3?

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In summary, the homework statement is to find the maximum and minimum of a function. The attempt at a solution is to find the maximum by finding the derivative with respect to x, y, and z and setting it to 0. The equation for the gradient vector is found and used to find the maximum.
  • #1
arcyqwerty
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Homework Statement


Find max/min of x^2+y^2+z^2 given x^4+y^4+z^4=3

Homework Equations


Use of gradient vectors related by LaGrange Multiplier

The Attempt at a Solution


[tex]\begin{gathered}
f\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2};g\left( {x,y,z} \right) = {x^4} + {y^4} + {z^4} - 3 = 0 \\
\vec \nabla f = \left\langle {2x,2y,2z} \right\rangle ;\vec \nabla g = \left\langle {4{x^3},4{y^3},4{z^3}} \right\rangle \\
\left\langle {2x,2y,2z} \right\rangle = \lambda \left\langle {4{x^3},4{y^3},4{z^3}} \right\rangle \\
2{x^2} = 2{y^2} = 2{z^2} \to x = \pm y = \pm z \\
3{x^4} - 3 = 0 \to {x^4} = 1 \to x = \pm 1 \to y = \pm 1,z = \pm 1 \\
\max = f\left( {1,1,1} \right) = f\left( {1,1, - 1} \right) = f\left( {1, - 1,1} \right) = f\left( {1, - 1, - 1} \right) = \\
f\left( { - 1,1,1} \right) = f\left( { - 1,1, - 1} \right) = f\left( { - 1, - 1,1} \right) = f\left( { - 1, - 1, - 1} \right) = 3 \\
\end{gathered}[/tex]

So I found the maximum but does the minimum exist?
 
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  • #2
arcyqwerty said:

Homework Statement


Find max/min of x^2+y^2+z^2 given x^4+y^4+z^4=3


Homework Equations


Use of gradient vectors related by LaGrange Multiplier


The Attempt at a Solution


[tex]\begin{gathered}
f\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2};g\left( {x,y,z} \right) = {x^4} + {y^4} + {z^4} - 3 = 0 \\
\vec \nabla f = \left\langle {2x,2y,2z} \right\rangle ;\vec \nabla g = \left\langle {4{x^3},4{y^3},4{z^3}} \right\rangle \\
\left\langle {2x,2y,2z} \right\rangle = \lambda \left\langle {4{x^3},4{y^3},4{z^3}} \right\rangle \\
2{x^2} = 2{y^2} = 2{z^2} \to x = \pm y = \pm z \\
3{x^4} - 3 = 0 \to {x^4} = 1 \to x = \pm 1 \to y = \pm 1,z = \pm 1 \\
\max = f\left( {1,1,1} \right) = f\left( {1,1, - 1} \right) = f\left( {1, - 1,1} \right) = f\left( {1, - 1, - 1} \right) = \\
f\left( { - 1,1,1} \right) = f\left( { - 1,1, - 1} \right) = f\left( { - 1, - 1,1} \right) = f\left( { - 1, - 1, - 1} \right) = 3 \\
\end{gathered}[/tex]

So I found the maximum but does the minimum exist?

Is the feasible set S = {(x,y,z): x^4 + y^4 + z^4 = 3} compact? Is the function f(x,y,z) = x^2 + y^2 + z^2 continuous on S? Have you heard of Weierstrass' Theorem?

RGV
 
  • #3
Ray Vickson said:
Is the feasible set S = {(x,y,z): x^4 + y^4 + z^4 = 3} compact? Is the function f(x,y,z) = x^2 + y^2 + z^2 continuous on S? Have you heard of Weierstrass' Theorem?

RGV

I'm not quite sure what you mean by compact or Weierstrass' Theorem but I think that the function is continuous
 
  • #4
arcyqwerty said:
I'm not quite sure what you mean by compact or Weierstrass' Theorem but I think that the function is continuous

Google is your friend.

RGV
 
  • #5
Ray Vickson said:
Google is your friend.

RGV

So...
"A subset S of a topological space X is compact if for every open cover of S there exists a finite subcover of S."

Not quite sure what that means exactly, but perhaps its compact if there can be a finite subset of the points defined by the function?

And...
There seems to be two different Theorems, one about estimating functions with polynomials and another about sequence convergence...
 
  • #6
arcyqwerty said:
So...
"A subset S of a topological space X is compact if for every open cover of S there exists a finite subcover of S."

Not quite sure what that means exactly, but perhaps its compact if there can be a finite subset of the points defined by the function?

And...
There seems to be two different Theorems, one about estimating functions with polynomials and another about sequence convergence...

If you keep searching you will eventually find a document in which all this is put into the context of ordinary 3-D space with the usual distance measure. In that case there is a theorem saying that a set is compact if and only if it is closed and bounded. So, is the set S closed (i.e., contains all its limit points)? Is it bounded? Then there is a theorem of Weierstrass saying that a continuous function on a compact set assumes both its maximum and its minimum. (These are theorems that are proven in advanced Calculus classes, well before 'topology'.) So, in your case the answer is YES: S is compact, and f has a minimum on S, as well as a maximum. None of this helps you *find* the minimum, but it does tell you that the search makes sense.

RGV
 

Related to Does a Minimum Exist for x^2+y^2+z^2 Given x^4+y^4+z^4=3?

1. What is the concept of "Max/min with constraints"?

"Max/min with constraints" is a mathematical optimization problem that involves finding the maximum or minimum value of a function, subject to certain constraints or limitations. These constraints can be in the form of equations or inequalities that restrict the values of the variables in the function.

2. How is "Max/min with constraints" different from unconstrained optimization?

In unconstrained optimization, there are no restrictions on the values of the variables in the function, so the goal is to simply find the maximum or minimum value. In "Max/min with constraints", the presence of constraints adds an additional layer of complexity as the optimal solution must satisfy both the objective function and the constraints.

3. What are some real-world applications of "Max/min with constraints"?

"Max/min with constraints" has various applications in fields such as economics, engineering, and operations research. For example, it can be used to determine the most cost-effective production levels for a company, the optimal design of a bridge, or the best route for a delivery truck given time and distance constraints.

4. What are some methods for solving "Max/min with constraints" problems?

There are several methods for solving "Max/min with constraints" problems, including the graphical method, the substitution method, the Lagrange multiplier method, and the simplex method. Each method has its own advantages and can be used depending on the complexity of the problem and the available resources.

5. How can "Max/min with constraints" be solved using computer algorithms?

Computer algorithms such as linear programming, quadratic programming, and nonlinear programming can be used to solve "Max/min with constraints" problems quickly and efficiently. These algorithms use mathematical techniques to find the optimal solution by iteratively testing different values for the variables until the objective function is maximized or minimized while satisfying the constraints.

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