Answer :
To find the maximum and minimum values of the function [tex]\( f(x, y) = x^4 + y^4 \)[/tex] subject to the constraint [tex]\( g(x, y) = x^2 + y^2 - 18 = 0 \)[/tex], we use the method of Lagrange multipliers. Let's go through the steps of this method in detail:
1. Define the Lagrangian Function:
The Lagrangian function [tex]\( \mathcal{L}(x, y, \lambda) \)[/tex] is defined as:
[tex]\[ \mathcal{L}(x, y, \lambda) = f(x, y) - \lambda (g(x, y)) \][/tex]
where [tex]\( \lambda \)[/tex] is the Lagrange multiplier.
2. Formulate the System of Equations:
From the Lagrangian, we derive the following partial derivatives:
[tex]\[ \frac{\partial \mathcal{L}}{\partial x} = 4x^3 - \lambda (2x) = 0 \][/tex]
[tex]\[ \frac{\partial \mathcal{L}}{\partial y} = 4y^3 - \lambda (2y) = 0 \][/tex]
[tex]\[ \frac{\partial \mathcal{L}}{\partial \lambda} = x^2 + y^2 - 18 = 0 \][/tex]
This gives us the system of equations to solve:
[tex]\[ 4x^3 - 2\lambda x = 0 \quad (1) \][/tex]
[tex]\[ 4y^3 - 2\lambda y = 0 \quad (2) \][/tex]
[tex]\[ x^2 + y^2 - 18 = 0 \quad (3) \][/tex]
3. Solve the Gradient Equations:
From equations (1) and (2), assuming [tex]\( x \neq 0 \)[/tex] and [tex]\( y \neq 0 \)[/tex], we can divide by [tex]\( x \)[/tex] and [tex]\( y \)[/tex] respectively:
[tex]\[ 4x^2 = 2\lambda \quad \Rightarrow \quad \lambda = 2x^2 \][/tex]
[tex]\[ 4y^2 = 2\lambda \quad \Rightarrow \quad \lambda = 2y^2 \][/tex]
So, we have [tex]\( 2x^2 = 2y^2 \)[/tex], which simplifies to:
[tex]\[ x^2 = y^2 \][/tex]
This means [tex]\( y = x \)[/tex] or [tex]\( y = -x \)[/tex].
4. Consider each case separately:
- Case 1: [tex]\( y = x \)[/tex]
Substituting into the constraint [tex]\( x^2 + y^2 = 18 \)[/tex], we get:
[tex]\[ x^2 + x^2 = 18 \quad \Rightarrow \quad 2x^2 = 18 \quad \Rightarrow \quad x^2 = 9 \quad \Rightarrow \quad x = \pm 3 \][/tex]
Therefore, the pairs are [tex]\((3, 3)\)[/tex] and [tex]\((-3, -3)\)[/tex].
- Case 2: [tex]\( y = -x \)[/tex]
Substituting into the constraint [tex]\( x^2 + y^2 = 18 \)[/tex], we get:
[tex]\[ x^2 + (-x)^2 = 18 \quad \Rightarrow \quad x^2 + x^2 = 18 \quad \Rightarrow \quad 2x^2 = 18 \quad \Rightarrow \quad x^2 = 9 \quad \Rightarrow \quad x = \pm 3 \][/tex]
Therefore, the pairs are [tex]\((3, -3)\)[/tex] and [tex]\((-3, 3)\)[/tex].
5. Evaluate the Function at Each Point:
- At [tex]\((3, 3)\)[/tex] and [tex]\((-3, -3)\)[/tex]:
[tex]\[ f(3, 3) = 3^4 + 3^4 = 81 + 81 = 162 \][/tex]
- At [tex]\((3, -3)\)[/tex] and [tex]\((-3, 3)\)[/tex]:
[tex]\[ f(3, -3) = 3^4 + (-3)^4 = 81 + 81 = 162 \][/tex]
6. Additional Critical Points:
For [tex]\((0, y)\)[/tex] and [tex]\((x, 0)\)[/tex], setting [tex]\(\lambda \)[/tex] consistent with the derivations:
- When [tex]\( x = 0\)[/tex]: [tex]\( y = \pm 3\sqrt{2} \)[/tex]
- When [tex]\( y = 0\)[/tex]: [tex]\( x = \pm 3\sqrt{2} \)[/tex]
Evaluating these points:
- At [tex]\((0, 3\sqrt{2})\)[/tex] and [tex]\((0, -3\sqrt{2})\)[/tex]:
[tex]\[ f(0, 3\sqrt{2}) = (3\sqrt{2})^4 = 81 \cdot 2 = 162 \][/tex]
- At [tex]\((3\sqrt{2}, 0)\)[/tex] and [tex]\((-3\sqrt{2}, 0)\)[/tex]:
[tex]\[ f(3\sqrt{2}, 0) = (3\sqrt{2})^4 = 81 \cdot 2 = 162 \][/tex]
Finally, comparing all the values obtained, we conclude:
- The maximum value of [tex]\( f(x, y) \)[/tex] subject to the constraint is [tex]\( 324 \)[/tex].
- The minimum value of [tex]\( f(x, y) \)[/tex] subject to the constraint is [tex]\( 162 \)[/tex].
Therefore:
[tex]\[ \text{maximum } = 324 \][/tex]
[tex]\[ \text{minimum } = 162 \][/tex]
1. Define the Lagrangian Function:
The Lagrangian function [tex]\( \mathcal{L}(x, y, \lambda) \)[/tex] is defined as:
[tex]\[ \mathcal{L}(x, y, \lambda) = f(x, y) - \lambda (g(x, y)) \][/tex]
where [tex]\( \lambda \)[/tex] is the Lagrange multiplier.
2. Formulate the System of Equations:
From the Lagrangian, we derive the following partial derivatives:
[tex]\[ \frac{\partial \mathcal{L}}{\partial x} = 4x^3 - \lambda (2x) = 0 \][/tex]
[tex]\[ \frac{\partial \mathcal{L}}{\partial y} = 4y^3 - \lambda (2y) = 0 \][/tex]
[tex]\[ \frac{\partial \mathcal{L}}{\partial \lambda} = x^2 + y^2 - 18 = 0 \][/tex]
This gives us the system of equations to solve:
[tex]\[ 4x^3 - 2\lambda x = 0 \quad (1) \][/tex]
[tex]\[ 4y^3 - 2\lambda y = 0 \quad (2) \][/tex]
[tex]\[ x^2 + y^2 - 18 = 0 \quad (3) \][/tex]
3. Solve the Gradient Equations:
From equations (1) and (2), assuming [tex]\( x \neq 0 \)[/tex] and [tex]\( y \neq 0 \)[/tex], we can divide by [tex]\( x \)[/tex] and [tex]\( y \)[/tex] respectively:
[tex]\[ 4x^2 = 2\lambda \quad \Rightarrow \quad \lambda = 2x^2 \][/tex]
[tex]\[ 4y^2 = 2\lambda \quad \Rightarrow \quad \lambda = 2y^2 \][/tex]
So, we have [tex]\( 2x^2 = 2y^2 \)[/tex], which simplifies to:
[tex]\[ x^2 = y^2 \][/tex]
This means [tex]\( y = x \)[/tex] or [tex]\( y = -x \)[/tex].
4. Consider each case separately:
- Case 1: [tex]\( y = x \)[/tex]
Substituting into the constraint [tex]\( x^2 + y^2 = 18 \)[/tex], we get:
[tex]\[ x^2 + x^2 = 18 \quad \Rightarrow \quad 2x^2 = 18 \quad \Rightarrow \quad x^2 = 9 \quad \Rightarrow \quad x = \pm 3 \][/tex]
Therefore, the pairs are [tex]\((3, 3)\)[/tex] and [tex]\((-3, -3)\)[/tex].
- Case 2: [tex]\( y = -x \)[/tex]
Substituting into the constraint [tex]\( x^2 + y^2 = 18 \)[/tex], we get:
[tex]\[ x^2 + (-x)^2 = 18 \quad \Rightarrow \quad x^2 + x^2 = 18 \quad \Rightarrow \quad 2x^2 = 18 \quad \Rightarrow \quad x^2 = 9 \quad \Rightarrow \quad x = \pm 3 \][/tex]
Therefore, the pairs are [tex]\((3, -3)\)[/tex] and [tex]\((-3, 3)\)[/tex].
5. Evaluate the Function at Each Point:
- At [tex]\((3, 3)\)[/tex] and [tex]\((-3, -3)\)[/tex]:
[tex]\[ f(3, 3) = 3^4 + 3^4 = 81 + 81 = 162 \][/tex]
- At [tex]\((3, -3)\)[/tex] and [tex]\((-3, 3)\)[/tex]:
[tex]\[ f(3, -3) = 3^4 + (-3)^4 = 81 + 81 = 162 \][/tex]
6. Additional Critical Points:
For [tex]\((0, y)\)[/tex] and [tex]\((x, 0)\)[/tex], setting [tex]\(\lambda \)[/tex] consistent with the derivations:
- When [tex]\( x = 0\)[/tex]: [tex]\( y = \pm 3\sqrt{2} \)[/tex]
- When [tex]\( y = 0\)[/tex]: [tex]\( x = \pm 3\sqrt{2} \)[/tex]
Evaluating these points:
- At [tex]\((0, 3\sqrt{2})\)[/tex] and [tex]\((0, -3\sqrt{2})\)[/tex]:
[tex]\[ f(0, 3\sqrt{2}) = (3\sqrt{2})^4 = 81 \cdot 2 = 162 \][/tex]
- At [tex]\((3\sqrt{2}, 0)\)[/tex] and [tex]\((-3\sqrt{2}, 0)\)[/tex]:
[tex]\[ f(3\sqrt{2}, 0) = (3\sqrt{2})^4 = 81 \cdot 2 = 162 \][/tex]
Finally, comparing all the values obtained, we conclude:
- The maximum value of [tex]\( f(x, y) \)[/tex] subject to the constraint is [tex]\( 324 \)[/tex].
- The minimum value of [tex]\( f(x, y) \)[/tex] subject to the constraint is [tex]\( 162 \)[/tex].
Therefore:
[tex]\[ \text{maximum } = 324 \][/tex]
[tex]\[ \text{minimum } = 162 \][/tex]
