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Magazine Chapter 13: Problem 6
Lagrange multipliers in two variables Use Lagrange multipliers to find the maximum and minimum values of \(f\) (when they exist) subject to the given constraint. $$f(x, y)=x y^{2} \text { subject to } x^{2}+y^{2}=1$$
Short Answer
Expert verified
Answer: The maximum value of the function is \(\frac{2}{3}\) and the minimum value is \(-\frac{2}{3}\).
Step by step solution
01
Identify the function and constraint
The function and constraint are given as:
$$f(x,y) = xy^2$$
$$g(x,y) = x^2 + y^2 = 1$$
02
Calculate gradients
Calculate the gradient of the function \(f\) and the constraint \(g\):
$$\nabla f(x, y) = \begin{bmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{bmatrix} = \begin{bmatrix} y^2 \\ 2xy \end{bmatrix}$$
$$\nabla g(x, y) = \begin{bmatrix} \frac{\partial g}{\partial x} \\ \frac{\partial g}{\partial y} \end{bmatrix} = \begin{bmatrix} 2x \\ 2y \end{bmatrix}$$
03
Set up the Lagrange multiplier equations
Introduce the Lagrange multiplier, \(\lambda\), and equate the gradients with a proportionality constant:
$$\nabla f(x, y) = \lambda \nabla g(x, y)$$
This gives us the following equations:
$$y^2 = 2x\lambda$$
$$2xy = 2y\lambda$$
04
Solve the system of equations
The system of equations we have is:
$$y^2 = 2x\lambda$$
$$2xy = 2y\lambda$$
$$x^2 + y^2 = 1$$
Now, we want to express \(\lambda\) in terms of \(x\) and \(y\) from the first two equations:
$$\lambda = \frac{y^2}{2x}$$ and
$$\lambda = \frac{xy}{y} = x$$
Since \(\lambda = x\), we have:
$$\frac{y^2}{2x} = x \implies y^2 = 2x^2$$
So, substitute the value of \(y^2\) into the constraint equation \(x^2 + y^2 = 1\):
$$x^2 + 2x^2 = 1 \implies x^2 = \frac{1}{3}$$
The \(x\) values can be \(x = \pm \sqrt{\frac{1}{3}}\). Now we can solve for \(y\). In the given constraint equation, replace \(x^2\) with \(\frac{1}{3}\):
$$\frac{1}{3} + y^2 = 1 \implies y^2 = \frac{2}{3}$$
The \(y\) values can be \(y = \pm \sqrt{\frac{2}{3}}\). This gives us four critical points:
$$(\sqrt{\frac{1}{3}}, \sqrt{\frac{2}{3}}),\ (-\sqrt{\frac{1}{3}}, \sqrt{\frac{2}{3}}),\ (\sqrt{\frac{1}{3}}, -\sqrt{\frac{2}{3}}),\ (-\sqrt{\frac{1}{3}}, -\sqrt{\frac{2}{3}})$$
05
Determine maximum and minimum values
Evaluate \(f(x,y)\) at each of the four critical points to find the maximum and minimum values:
$$f(\sqrt{\frac{1}{3}}, \sqrt{\frac{2}{3}}) = \left(\sqrt{\frac{1}{3}}\right)\left(\sqrt{\frac{2}{3}}\right)^2 = \frac{2}{3}$$
$$f(-\sqrt{\frac{1}{3}}, \sqrt{\frac{2}{3}}) = -\frac{2}{3}$$
$$f(\sqrt{\frac{1}{3}}, -\sqrt{\frac{2}{3}}) = -\frac{2}{3}$$
$$f(-\sqrt{\frac{1}{3}}, -\sqrt{\frac{2}{3}}) = \frac{2}{3}$$
From the result, the maximum value of \(f(x,y)\) is \(\frac{2}{3}\) and the minimum value is \(-\frac{2}{3}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Critical Points in Lagrange Multipliers
Critical points represent the values in the domain of a function where its partial derivatives are zero or undefined. They are essential in optimization problems because they allow us to find potential locations of maxima and minima. When applying Lagrange multipliers, critical points play a crucial role.
In the given exercise, our goal is to locate the critical points of the function subject to a constraint. The given function is \( f(x, y) = xy^2 \) and the constraint is \( g(x, y) = x^2 + y^2 = 1 \). To apply the method of Lagrange multipliers, we calculate the gradients of both the function and the constraint.
\[ abla f(x, y) = \begin{bmatrix} y^2 \ 2xy \end{bmatrix} \]\[ abla g(x, y) = \begin{bmatrix} 2x \ 2y \end{bmatrix} \]By setting these two gradients proportional through a Lagrange multiplier \( \lambda \),
we subsequently generate equations that allow us to identify critical points within the constraint's boundary.
In the given exercise, our goal is to locate the critical points of the function subject to a constraint. The given function is \( f(x, y) = xy^2 \) and the constraint is \( g(x, y) = x^2 + y^2 = 1 \). To apply the method of Lagrange multipliers, we calculate the gradients of both the function and the constraint.
\[ abla f(x, y) = \begin{bmatrix} y^2 \ 2xy \end{bmatrix} \]\[ abla g(x, y) = \begin{bmatrix} 2x \ 2y \end{bmatrix} \]By setting these two gradients proportional through a Lagrange multiplier \( \lambda \),
we subsequently generate equations that allow us to identify critical points within the constraint's boundary.
Optimization through Lagrange Multipliers
Optimization in mathematics focuses on finding the best solution in a range of possible solutions, which could involve seeking a maximum or minimum value of a function. The method of Lagrange multipliers is a powerful tool in solving optimization problems that involve constraints.
The exercise demonstrates optimization by using Lagrange multipliers to either maximize or minimize the function \( f(x, y) = xy^2 \), subject to \( x^2 + y^2 = 1 \), a circular boundary. We start by equating the gradients of the function and the constraint multiplied by \( \lambda \):
The exercise demonstrates optimization by using Lagrange multipliers to either maximize or minimize the function \( f(x, y) = xy^2 \), subject to \( x^2 + y^2 = 1 \), a circular boundary. We start by equating the gradients of the function and the constraint multiplied by \( \lambda \):
- \( y^2 = 2x\lambda \)
- \( 2xy = 2y\lambda \)
Using the Gradient Method
The gradient method is a straightforward optimization approach that seeks directionally based maxima and minima of functions. In the context of Lagrange multipliers, it aligns these gradients in a way that respects constraints, offering a meticulously crafted path.
When using gradients:
Ultimately, the gradient method uses these directionally rich insights, married via Lagrange multipliers, to explore function behavior under constraints, efficiently identifying points of interest for optimization.
When using gradients:
- The function's gradient, \( abla f(x, y) = \begin{bmatrix} y^2 \ 2xy \end{bmatrix} \), highlights the steepest ascent in unconstrained space.
- Additionally, \( abla g(x, y) = \begin{bmatrix} 2x \ 2y \end{bmatrix} \), reflecting the constraint's direction.
Ultimately, the gradient method uses these directionally rich insights, married via Lagrange multipliers, to explore function behavior under constraints, efficiently identifying points of interest for optimization.
