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To find the maximum and minimum values of the function [tex]\( f(x, y) = x^2 + 5y \)[/tex] subject to the constraint [tex]\( 4x + y = 9 \)[/tex], we can use the method of Lagrange multipliers. Here is a detailed, step-by-step solution.
1. Define the Lagrangian function:
The Lagrangian function is given by
[tex]\[ \mathcal{L}(x, y, \lambda) = f(x, y) - \lambda (4x + y - 9) \][/tex]
where [tex]\(\lambda\)[/tex] is the Lagrange multiplier.
2. Formulate the Lagrangian:
Substitute [tex]\( f(x, y) \)[/tex] and the constraint into the Lagrangian function:
[tex]\[ \mathcal{L}(x, y, \lambda) = x^2 + 5y - \lambda (4x + y - 9) \][/tex]
3. Compute the partial derivatives:
Compute the partial derivatives of [tex]\(\mathcal{L}\)[/tex] with respect to [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(\lambda\)[/tex], and set them to zero:
[tex]\[ \frac{\partial \mathcal{L}}{\partial x} = 2x - 4\lambda = 0 \][/tex]
[tex]\[ \frac{\partial \mathcal{L}}{\partial y} = 5 - \lambda = 0 \][/tex]
[tex]\[ \frac{\partial \mathcal{L}}{\partial \lambda} = -(4x + y - 9) = 0 \][/tex]
4. Solve the system of equations:
From [tex]\( \frac{\partial \mathcal{L}}{\partial x} = 2x - 4\lambda = 0 \)[/tex], we get:
[tex]\[ \lambda = \frac{x}{2} \][/tex]
From [tex]\( \frac{\partial \mathcal{L}}{\partial y} = 5 - \lambda = 0 \)[/tex], we get:
[tex]\[ \lambda = 5 \][/tex]
Equating the two expressions for [tex]\(\lambda\)[/tex] gives us:
[tex]\[ \frac{x}{2} = 5 \implies x = 10 \][/tex]
Substituting [tex]\(x = 10\)[/tex] into the constraint [tex]\( 4x + y = 9 \)[/tex] gives:
[tex]\[ 4(10) + y = 9 \implies 40 + y = 9 \implies y = 9 - 40 \implies y = -31 \][/tex]
5. Substitute the values into the objective function:
Substituting [tex]\(x = 10\)[/tex] and [tex]\(y = -31\)[/tex] into [tex]\( f(x, y) = x^2 + 5y \)[/tex]:
[tex]\[ f(10, -31) = 10^2 + 5(-31) = 100 - 155 = -55 \][/tex]
Thus, the function has a single critical point at [tex]\((10, -31)\)[/tex]. Evaluating [tex]\(f\)[/tex] at this point, we get [tex]\(-55\)[/tex].
Since there are no other points to check within the domain defined by the constraint, both the maximum and the minimum values are [tex]\(-55\)[/tex].
Therefore, the maximum value is [tex]\(\boxed{-55}\)[/tex] and the minimum value is [tex]\(\boxed{-55}\)[/tex].
1. Define the Lagrangian function:
The Lagrangian function is given by
[tex]\[ \mathcal{L}(x, y, \lambda) = f(x, y) - \lambda (4x + y - 9) \][/tex]
where [tex]\(\lambda\)[/tex] is the Lagrange multiplier.
2. Formulate the Lagrangian:
Substitute [tex]\( f(x, y) \)[/tex] and the constraint into the Lagrangian function:
[tex]\[ \mathcal{L}(x, y, \lambda) = x^2 + 5y - \lambda (4x + y - 9) \][/tex]
3. Compute the partial derivatives:
Compute the partial derivatives of [tex]\(\mathcal{L}\)[/tex] with respect to [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(\lambda\)[/tex], and set them to zero:
[tex]\[ \frac{\partial \mathcal{L}}{\partial x} = 2x - 4\lambda = 0 \][/tex]
[tex]\[ \frac{\partial \mathcal{L}}{\partial y} = 5 - \lambda = 0 \][/tex]
[tex]\[ \frac{\partial \mathcal{L}}{\partial \lambda} = -(4x + y - 9) = 0 \][/tex]
4. Solve the system of equations:
From [tex]\( \frac{\partial \mathcal{L}}{\partial x} = 2x - 4\lambda = 0 \)[/tex], we get:
[tex]\[ \lambda = \frac{x}{2} \][/tex]
From [tex]\( \frac{\partial \mathcal{L}}{\partial y} = 5 - \lambda = 0 \)[/tex], we get:
[tex]\[ \lambda = 5 \][/tex]
Equating the two expressions for [tex]\(\lambda\)[/tex] gives us:
[tex]\[ \frac{x}{2} = 5 \implies x = 10 \][/tex]
Substituting [tex]\(x = 10\)[/tex] into the constraint [tex]\( 4x + y = 9 \)[/tex] gives:
[tex]\[ 4(10) + y = 9 \implies 40 + y = 9 \implies y = 9 - 40 \implies y = -31 \][/tex]
5. Substitute the values into the objective function:
Substituting [tex]\(x = 10\)[/tex] and [tex]\(y = -31\)[/tex] into [tex]\( f(x, y) = x^2 + 5y \)[/tex]:
[tex]\[ f(10, -31) = 10^2 + 5(-31) = 100 - 155 = -55 \][/tex]
Thus, the function has a single critical point at [tex]\((10, -31)\)[/tex]. Evaluating [tex]\(f\)[/tex] at this point, we get [tex]\(-55\)[/tex].
Since there are no other points to check within the domain defined by the constraint, both the maximum and the minimum values are [tex]\(-55\)[/tex].
Therefore, the maximum value is [tex]\(\boxed{-55}\)[/tex] and the minimum value is [tex]\(\boxed{-55}\)[/tex].
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