Math Problem Statement

find the maximum value of the function f(x,y,z) = x+2z under the two constraints of (x^2)+2(y^2)=10 and (x^2)+(y^2)+4(z^2)=8

Solution

We are tasked with finding the maximum value of the function f(x,y,z)=x+2zf(x, y, z) = x + 2z under the constraints:

g1(x,y)=x2+2y2=10g_1(x, y) = x^2 + 2y^2 = 10 g2(x,y,z)=x2+y2+4z2=8g_2(x, y, z) = x^2 + y^2 + 4z^2 = 8

Step 1: Lagrange Multipliers

To solve this, we apply the method of Lagrange multipliers. We need to find f\nabla f, g1\nabla g_1, and g2\nabla g_2, where \nabla denotes the gradient.

  1. f(x,y,z)=(1,0,2)\nabla f(x, y, z) = \left( 1, 0, 2 \right)
  2. g1(x,y)=(2x,4y,0)\nabla g_1(x, y) = \left( 2x, 4y, 0 \right)
  3. g2(x,y,z)=(2x,2y,8z)\nabla g_2(x, y, z) = \left( 2x, 2y, 8z \right)

We now set up the Lagrange multiplier equations:

f=λg1+μg2\nabla f = \lambda \nabla g_1 + \mu \nabla g_2

This gives the system of equations:

  1. 1=λ(2x)+μ(2x)1 = \lambda (2x) + \mu (2x)
  2. 0=λ(4y)+μ(2y)0 = \lambda (4y) + \mu (2y)
  3. 2=λ(0)+μ(8z)2 = \lambda (0) + \mu (8z)

Step 2: Solving the system

From the second equation:

0=λ(4y)+μ(2y)0 = \lambda (4y) + \mu (2y)

For this to hold, either y=0y = 0 or λ(4)+μ(2)=0\lambda (4) + \mu (2) = 0. Let's consider both cases.

Case 1: y=0y = 0

If y=0y = 0, the constraints become:

  1. x2=10x^2 = 10, so x=±10x = \pm \sqrt{10}
  2. x2+4z2=8x^2 + 4z^2 = 8, substituting x2=10x^2 = 10:

10+4z2=84z2=210 + 4z^2 = 8 \quad \Rightarrow \quad 4z^2 = -2

This yields no real solution for zz. Therefore, y=0y = 0 does not lead to a valid solution.

Case 2: λ(4)+μ(2)=0\lambda (4) + \mu (2) = 0

This implies:

λ=μ2\lambda = -\frac{\mu}{2}

Substitute this into the other two equations:

  1. 1=λ(2x)+μ(2x)1=(μ2)(2x)+μ(2x)=μx1 = \lambda (2x) + \mu (2x) \Rightarrow 1 = (-\frac{\mu}{2})(2x) + \mu(2x) = \mu x
  2. 2=μ(8z)μ=14z2 = \mu(8z) \Rightarrow \mu = \frac{1}{4z}

Substitute μ=14z\mu = \frac{1}{4z} into 1=μx1 = \mu x:

1=x4zx=4z1 = \frac{x}{4z} \quad \Rightarrow \quad x = 4z

Step 3: Substituting into the constraints

Now, substitute x=4zx = 4z into the constraints:

  1. (4z)2+2y2=1016z2+2y2=102y2=1016z2y2=58z2(4z)^2 + 2y^2 = 10 \Rightarrow 16z^2 + 2y^2 = 10 \Rightarrow 2y^2 = 10 - 16z^2 \Rightarrow y^2 = 5 - 8z^2
  2. (4z)2+y2+4z2=816z2+y2+4z2=820z2+y2=8(4z)^2 + y^2 + 4z^2 = 8 \Rightarrow 16z^2 + y^2 + 4z^2 = 8 \Rightarrow 20z^2 + y^2 = 8

Substitute y2=58z2y^2 = 5 - 8z^2 into this equation:

20z2+(58z2)=812z2+5=812z2=3z2=1420z^2 + (5 - 8z^2) = 8 \quad \Rightarrow \quad 12z^2 + 5 = 8 \quad \Rightarrow \quad 12z^2 = 3 \quad \Rightarrow \quad z^2 = \frac{1}{4}

So, z=±12z = \pm \frac{1}{2}.

Step 4: Find corresponding values for xx and yy

If z=12z = \frac{1}{2}, then x=4z=2x = 4z = 2.

Substitute z=12z = \frac{1}{2} into y2=58z2y^2 = 5 - 8z^2:

y2=58(12)2=52=3y=±3y^2 = 5 - 8\left(\frac{1}{2}\right)^2 = 5 - 2 = 3 \quad \Rightarrow \quad y = \pm \sqrt{3}

Step 5: Maximum value of f(x,y,z)f(x, y, z)

The function is f(x,y,z)=x+2zf(x, y, z) = x + 2z. Substituting x=2x = 2 and z=12z = \frac{1}{2}:

f(x,y,z)=2+2(12)=3f(x, y, z) = 2 + 2\left(\frac{1}{2}\right) = 3

Thus, the maximum value of f(x,y,z)=3f(x, y, z) = 3.

Would you like further details or clarifications? Here are some related questions to explore:

  1. How does the method of Lagrange multipliers work in multivariable optimization?
  2. What are the geometric interpretations of constraints in optimization problems?
  3. Can the method of Lagrange multipliers solve problems with inequality constraints?
  4. How does the solution change if the constraints are relaxed or modified?
  5. What are the conditions for a local maximum or minimum in constrained optimization?

Tip: Always check for boundary conditions or special cases, like zero values, when solving optimization problems.

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Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Optimization
Lagrange Multipliers

Formulas

Lagrange Multiplier Equation: ∇f = λ∇g₁ + μ∇g₂
f(x, y, z) = x + 2z
g₁(x, y) = x^2 + 2y^2 = 10
g₂(x, y, z) = x^2 + y^2 + 4z^2 = 8

Theorems

Lagrange Multiplier Theorem

Suitable Grade Level

Undergraduate (Calculus III or Advanced Mathematics)