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To solve the problem of finding the maximum and minimum values of the function [tex]\( f(x, y) = 8xy \)[/tex] subject to the constraint [tex]\( x^2 + y^2 = 9 \)[/tex], we can use the method of Lagrange multipliers.
### Step 1: Formulate the Lagrange Function
First, define the constraint [tex]\( g(x, y) = x^2 + y^2 - 9 = 0 \)[/tex]. According to the method of Lagrange multipliers, we introduce a new variable [tex]\( \lambda \)[/tex] and form the Lagrange function [tex]\( \mathcal{L} \)[/tex]:
[tex]\[ \mathcal{L}(x, y, \lambda) = f(x, y) + \lambda g(x, y) \][/tex]
Substitute [tex]\( f(x, y) \)[/tex] and [tex]\( g(x, y) \)[/tex] into the Lagrange function:
[tex]\[ \mathcal{L}(x, y, \lambda) = 8xy + \lambda (x^2 + y^2 - 9) \][/tex]
### Step 2: Take Partial Derivatives
Compute the partial derivatives of [tex]\( \mathcal{L} \)[/tex] with respect to [tex]\( x \)[/tex], [tex]\( y \)[/tex], and [tex]\( \lambda \)[/tex]:
[tex]\[ \frac{\partial \mathcal{L}}{\partial x} = 8y + \lambda 2x \][/tex]
[tex]\[ \frac{\partial \mathcal{L}}{\partial y} = 8x + \lambda 2y \][/tex]
[tex]\[ \frac{\partial \mathcal{L}}{\partial \lambda} = x^2 + y^2 - 9 \][/tex]
### Step 3: Solve the System of Equations
Set the partial derivatives equal to zero and solve the system of equations:
[tex]\[ 8y + \lambda 2x = 0 \][/tex]
[tex]\[ 8x + \lambda 2y = 0 \][/tex]
[tex]\[ x^2 + y^2 - 9 = 0 \][/tex]
From the first two equations, we get:
[tex]\[ \lambda = -\frac{8y}{2x} = -\frac{4y}{x} \][/tex]
[tex]\[ \lambda = -\frac{8x}{2y} = -\frac{4x}{y} \][/tex]
Since both expressions for [tex]\( \lambda \)[/tex] must be equal, set them equal to each other:
[tex]\[ -\frac{4y}{x} = -\frac{4x}{y} \][/tex]
[tex]\[ \frac{y}{x} = \frac{x}{y} \][/tex]
This implies [tex]\( y^2 = x^2 \)[/tex], so [tex]\( y = x \)[/tex] or [tex]\( y = -x \)[/tex].
### Step 4: Substitute Back into the Constraint
Substitute [tex]\( y = x \)[/tex] and [tex]\( y = -x \)[/tex] into the constraint [tex]\( x^2 + y^2 = 9 \)[/tex]:
1. For [tex]\( y = x \)[/tex]:
[tex]\[ x^2 + x^2 = 9 \][/tex]
[tex]\[ 2x^2 = 9 \][/tex]
[tex]\[ x^2 = \frac{9}{2} \][/tex]
[tex]\[ x = \pm \frac{3}{\sqrt{2}} \][/tex]
So, [tex]\( y = \pm \frac{3}{\sqrt{2}} \)[/tex].
2. For [tex]\( y = -x \)[/tex]:
[tex]\[ x^2 + (-x)^2 = 9 \][/tex]
[tex]\[ 2x^2 = 9 \][/tex]
[tex]\[ x^2 = \frac{9}{2} \][/tex]
[tex]\[ x = \pm \frac{3}{\sqrt{2}} \][/tex]
So, [tex]\( y = \mp \frac{3}{\sqrt{2}} \)[/tex].
### Step 5: Evaluate [tex]\( f(x, y) \)[/tex]
Evaluate [tex]\( f(x, y) = 8xy \)[/tex] at the critical points:
1. For [tex]\( (x, y) = \left(\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right) \)[/tex] and [tex]\( \left(-\frac{3}{\sqrt{2}}, -\frac{3}{\sqrt{2}}\right) \)[/tex]:
[tex]\[ f\left(\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right) = 8 \cdot \frac{3}{\sqrt{2}} \cdot \frac{3}{\sqrt{2}} = 8 \cdot \frac{9}{2} = 36 \][/tex]
[tex]\[ f\left(-\frac{3}{\sqrt{2}}, -\frac{3}{\sqrt{2}}\right) = 36 \][/tex]
2. For [tex]\( (x, y) = \left(\frac{3}{\sqrt{2}}, -\frac{3}{\sqrt{2}}\right) \)[/tex] and [tex]\( \left(-\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right) \)[/tex]:
[tex]\[ f\left(\frac{3}{\sqrt{2}}, -\frac{3}{\sqrt{2}}\right) = 8 \cdot \frac{3}{\sqrt{2}} \cdot -\frac{3}{\sqrt{2}} = 8 \cdot -\frac{9}{2} = -36 \][/tex]
[tex]\[ f\left(-\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right) = -36 \][/tex]
### Conclusion
The maximum value of [tex]\( f(x, y) \)[/tex] is [tex]\( 36 \)[/tex].
The minimum value of [tex]\( f(x, y) \)[/tex] is [tex]\( -36 \)[/tex].
So the answers are:
[tex]\[ \text{maximum } = 36 \][/tex]
[tex]\[ \text{minimum } = -36 \][/tex]
### Step 1: Formulate the Lagrange Function
First, define the constraint [tex]\( g(x, y) = x^2 + y^2 - 9 = 0 \)[/tex]. According to the method of Lagrange multipliers, we introduce a new variable [tex]\( \lambda \)[/tex] and form the Lagrange function [tex]\( \mathcal{L} \)[/tex]:
[tex]\[ \mathcal{L}(x, y, \lambda) = f(x, y) + \lambda g(x, y) \][/tex]
Substitute [tex]\( f(x, y) \)[/tex] and [tex]\( g(x, y) \)[/tex] into the Lagrange function:
[tex]\[ \mathcal{L}(x, y, \lambda) = 8xy + \lambda (x^2 + y^2 - 9) \][/tex]
### Step 2: Take Partial Derivatives
Compute the partial derivatives of [tex]\( \mathcal{L} \)[/tex] with respect to [tex]\( x \)[/tex], [tex]\( y \)[/tex], and [tex]\( \lambda \)[/tex]:
[tex]\[ \frac{\partial \mathcal{L}}{\partial x} = 8y + \lambda 2x \][/tex]
[tex]\[ \frac{\partial \mathcal{L}}{\partial y} = 8x + \lambda 2y \][/tex]
[tex]\[ \frac{\partial \mathcal{L}}{\partial \lambda} = x^2 + y^2 - 9 \][/tex]
### Step 3: Solve the System of Equations
Set the partial derivatives equal to zero and solve the system of equations:
[tex]\[ 8y + \lambda 2x = 0 \][/tex]
[tex]\[ 8x + \lambda 2y = 0 \][/tex]
[tex]\[ x^2 + y^2 - 9 = 0 \][/tex]
From the first two equations, we get:
[tex]\[ \lambda = -\frac{8y}{2x} = -\frac{4y}{x} \][/tex]
[tex]\[ \lambda = -\frac{8x}{2y} = -\frac{4x}{y} \][/tex]
Since both expressions for [tex]\( \lambda \)[/tex] must be equal, set them equal to each other:
[tex]\[ -\frac{4y}{x} = -\frac{4x}{y} \][/tex]
[tex]\[ \frac{y}{x} = \frac{x}{y} \][/tex]
This implies [tex]\( y^2 = x^2 \)[/tex], so [tex]\( y = x \)[/tex] or [tex]\( y = -x \)[/tex].
### Step 4: Substitute Back into the Constraint
Substitute [tex]\( y = x \)[/tex] and [tex]\( y = -x \)[/tex] into the constraint [tex]\( x^2 + y^2 = 9 \)[/tex]:
1. For [tex]\( y = x \)[/tex]:
[tex]\[ x^2 + x^2 = 9 \][/tex]
[tex]\[ 2x^2 = 9 \][/tex]
[tex]\[ x^2 = \frac{9}{2} \][/tex]
[tex]\[ x = \pm \frac{3}{\sqrt{2}} \][/tex]
So, [tex]\( y = \pm \frac{3}{\sqrt{2}} \)[/tex].
2. For [tex]\( y = -x \)[/tex]:
[tex]\[ x^2 + (-x)^2 = 9 \][/tex]
[tex]\[ 2x^2 = 9 \][/tex]
[tex]\[ x^2 = \frac{9}{2} \][/tex]
[tex]\[ x = \pm \frac{3}{\sqrt{2}} \][/tex]
So, [tex]\( y = \mp \frac{3}{\sqrt{2}} \)[/tex].
### Step 5: Evaluate [tex]\( f(x, y) \)[/tex]
Evaluate [tex]\( f(x, y) = 8xy \)[/tex] at the critical points:
1. For [tex]\( (x, y) = \left(\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right) \)[/tex] and [tex]\( \left(-\frac{3}{\sqrt{2}}, -\frac{3}{\sqrt{2}}\right) \)[/tex]:
[tex]\[ f\left(\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right) = 8 \cdot \frac{3}{\sqrt{2}} \cdot \frac{3}{\sqrt{2}} = 8 \cdot \frac{9}{2} = 36 \][/tex]
[tex]\[ f\left(-\frac{3}{\sqrt{2}}, -\frac{3}{\sqrt{2}}\right) = 36 \][/tex]
2. For [tex]\( (x, y) = \left(\frac{3}{\sqrt{2}}, -\frac{3}{\sqrt{2}}\right) \)[/tex] and [tex]\( \left(-\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right) \)[/tex]:
[tex]\[ f\left(\frac{3}{\sqrt{2}}, -\frac{3}{\sqrt{2}}\right) = 8 \cdot \frac{3}{\sqrt{2}} \cdot -\frac{3}{\sqrt{2}} = 8 \cdot -\frac{9}{2} = -36 \][/tex]
[tex]\[ f\left(-\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right) = -36 \][/tex]
### Conclusion
The maximum value of [tex]\( f(x, y) \)[/tex] is [tex]\( 36 \)[/tex].
The minimum value of [tex]\( f(x, y) \)[/tex] is [tex]\( -36 \)[/tex].
So the answers are:
[tex]\[ \text{maximum } = 36 \][/tex]
[tex]\[ \text{minimum } = -36 \][/tex]
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