Math Problem Statement

Find the maximum and minimum values of x + y + z subject to the constraint \(\frac{x^2}{4} + \frac{y^2}{5} + \frac{z^2}{25} = 1\).

Solution

The problem is asking to find the maximum and minimum values of x+y+zx + y + z, subject to the constraint:

x24+y25+z225=1.\frac{x^2}{4} + \frac{y^2}{5} + \frac{z^2}{25} = 1.

This can be solved using the method of Lagrange multipliers. Here's the general approach:

Step 1: Set up the Lagrange function

Let f(x,y,z)=x+y+zf(x, y, z) = x + y + z be the function to be maximized or minimized, and the constraint function is:

g(x,y,z)=x24+y25+z2251=0.g(x, y, z) = \frac{x^2}{4} + \frac{y^2}{5} + \frac{z^2}{25} - 1 = 0.

The Lagrangian function is defined as:

L(x,y,z,λ)=f(x,y,z)λg(x,y,z).\mathcal{L}(x, y, z, \lambda) = f(x, y, z) - \lambda g(x, y, z).

This becomes:

L(x,y,z,λ)=(x+y+z)λ(x24+y25+z2251).\mathcal{L}(x, y, z, \lambda) = (x + y + z) - \lambda \left( \frac{x^2}{4} + \frac{y^2}{5} + \frac{z^2}{25} - 1 \right).

Step 2: Compute partial derivatives

We need to find the critical points by setting the partial derivatives of the Lagrange function with respect to xx, yy, zz, and λ\lambda to zero.

  • Lx=1λx2=0\frac{\partial \mathcal{L}}{\partial x} = 1 - \lambda \cdot \frac{x}{2} = 0
  • Ly=1λ2y5=0\frac{\partial \mathcal{L}}{\partial y} = 1 - \lambda \cdot \frac{2y}{5} = 0
  • Lz=1λ2z25=0\frac{\partial \mathcal{L}}{\partial z} = 1 - \lambda \cdot \frac{2z}{25} = 0
  • Lλ=x24+y25+z2251=0\frac{\partial \mathcal{L}}{\partial \lambda} = \frac{x^2}{4} + \frac{y^2}{5} + \frac{z^2}{25} - 1 = 0.

Step 3: Solve the system of equations

We now solve the system of equations obtained from the partial derivatives. Let me work through the calculations.The solutions to the system of equations are:

  1. x=23417,y=53434,z=253434x = -\frac{2\sqrt{34}}{17}, y = -\frac{5\sqrt{34}}{34}, z = -\frac{25\sqrt{34}}{34}
  2. x=23417,y=53434,z=253434x = \frac{2\sqrt{34}}{17}, y = \frac{5\sqrt{34}}{34}, z = \frac{25\sqrt{34}}{34}

Step 4: Calculate x+y+zx + y + z for each solution

  • For the first solution: x+y+z=2341753434253434x + y + z = -\frac{2\sqrt{34}}{17} - \frac{5\sqrt{34}}{34} - \frac{25\sqrt{34}}{34}
  • For the second solution: x+y+z=23417+53434+253434x + y + z = \frac{2\sqrt{34}}{17} + \frac{5\sqrt{34}}{34} + \frac{25\sqrt{34}}{34}

Let me calculate these values.The values of x+y+zx + y + z are:

  • Minimum value: 5.83-5.83
  • Maximum value: 5.835.83

Thus, the maximum value of x+y+zx + y + z is approximately 5.835.83, and the minimum value is approximately 5.83-5.83, subject to the given constraint.

Would you like further details or clarifications on any part of the solution?

Here are 5 related questions you can explore:

  1. What is the geometric interpretation of using Lagrange multipliers?
  2. How does the method of Lagrange multipliers work in cases with more constraints?
  3. Can the maximum and minimum values of x+y+zx + y + z be calculated using different methods?
  4. How would the problem change if the constraint were different (e.g., an inequality)?
  5. What happens to the optimization problem if one of the variables, say zz, is held constant?

Tip: When solving optimization problems with constraints, always ensure the constraint is active (equality holds) at the critical points to avoid missing solutions.

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Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Lagrange Multipliers
Optimization

Formulas

Lagrange function: \( \mathcal{L}(x, y, z, \lambda) = f(x, y, z) - \lambda \cdot g(x, y, z) \)
Constraint equation: \(\frac{x^2}{4} + \frac{y^2}{5} + \frac{z^2}{25} = 1\)

Theorems

Method of Lagrange Multipliers

Suitable Grade Level

Undergraduate Calculus