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Magazine Chapter 10: Problem 7
Find the maximum and minimum values of \(f(x, y)=x y\) on the ellipse \(5 x^{2}+y^{2}=3\) maximum value \(=\) __________. minimum value \(=\) __________.
Short Answer
Expert verified
maximum value \(= \frac{1}{\sqrt{5}}\).
minimum value \(= -\frac{1}{\sqrt{5}}\).
Step by step solution
01
Rewrite the constraint and the function
Our constraint is the equation of the ellipse, which is given by:
\[5x^2 + y^2 = 3\]
Let's rewrite this equation as:
\[g(x, y) = 5x^2 + y^2 - 3\]
Our function that we want to maximize and minimize is:
\[f(x, y) = xy\]
02
Apply the method of Lagrange multipliers
To find the maximum and minimum values of \(f(x, y)\) on the ellipse, we use the method of Lagrange multipliers. We need to find the gradient of the function and the gradient of the constraint and set them proportional to each other:
\[\nabla f(x, y) = \lambda \nabla g(x, y)\]
Calculate the gradient of \(f(x, y)\) and \(g(x, y)\):
\[\nabla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right) = (y, x)\]
\[\nabla g(x, y) = \left(\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}\right) = (10x, 2y)\]
Now set the gradient of \(f(x, y)\) proportional to the gradient of \(g(x, y)\) using a constant \(\lambda\):
\[(y, x) = \lambda(10x, 2y)\]
03
Solve the equations for critical points
We now have the following system of equations:
\[\begin{cases}
y = 10\lambda x \\
x = 2\lambda y \\
5x^2 + y^2 - 3 = 0
\end{cases}\]
First, we can solve for \(\lambda\):
\[\lambda = \frac{y}{10x}\]
\[\lambda = \frac{x}{2y}\]
By equating the two expressions for \(\lambda\), we get:
\[\frac{y}{10x} = \frac{x}{2y}\]
Solve for \(y\):
\[y^2 = \frac{x^2}{5}\]
Now substitute this equation back into the constraint equation (ellipse equation):
\[5x^2 + \frac{x^2}{5} - 3 = 0\]
Solve for \(x\):
\[x^2 = 1\]
This gives us two critical points:
\[x = \pm 1\]
04
Find the corresponding y-values for the critical points
Now find the corresponding y-values for the critical points by substituting the x-values back into the equation \(y^2 = \frac{x^2}{5}\):
\[y^2 = \frac{1}{5}\]
This gives us two critical points:
\[y = \pm \frac{1}{\sqrt{5}}\]
So we have four critical points: \((1, \frac{1}{\sqrt{5}}), (1, -\frac{1}{\sqrt{5}}), (-1, \frac{1}{\sqrt{5}}), (-1, -\frac{1}{\sqrt{5}})\)
05
Evaluate the function at the critical points
Now evaluate the function \(f(x, y) = xy\) at each of the critical points:
\[f(1, \frac{1}{\sqrt{5}}) = \frac{1}{\sqrt{5}}\]
\[f(1, -\frac{1}{\sqrt{5}}) = -\frac{1}{\sqrt{5}}\]
\[f(-1, \frac{1}{\sqrt{5}}) = -\frac{1}{\sqrt{5}}\]
\[f(-1, -\frac{1}{\sqrt{5}}) = \frac{1}{\sqrt{5}}\]
From this, we can see that the maximum value is \(\frac{1}{\sqrt{5}}\) and the minimum value is \(-\frac{1}{\sqrt{5}}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Critical Points
In the context of optimization using Lagrange multipliers, critical points are essential as these are the points at which the function might reach a local maximum, local minimum, or a saddle point.
One important aspect to understand about critical points is how they are determined. For a given function constrained by an equation (like an ellipse), critical points are found where the gradients of the function and the constraint are proportional. This is accomplished by setting up a system of equations based on the Lagrange multipliers r> After obtaining these equations, solving them reveals potential points on the constraint that are critical, meaning the function's values might change behavior at these points.
In our exercise, critical points were found by solving the system of equations derived from r> \[ y = 10\lambda x \]\[ x = 2\lambda y \]\[ 5x^2 + y^2 - 3 = 0 \]
These relate back to identifying points on the constraint, the ellipse, where the function could have a minimum or maximum.
One important aspect to understand about critical points is how they are determined. For a given function constrained by an equation (like an ellipse), critical points are found where the gradients of the function and the constraint are proportional. This is accomplished by setting up a system of equations based on the Lagrange multipliers r> After obtaining these equations, solving them reveals potential points on the constraint that are critical, meaning the function's values might change behavior at these points.
In our exercise, critical points were found by solving the system of equations derived from r> \[ y = 10\lambda x \]\[ x = 2\lambda y \]\[ 5x^2 + y^2 - 3 = 0 \]
These relate back to identifying points on the constraint, the ellipse, where the function could have a minimum or maximum.
Optimization
Optimization involving multivariable functions, like those in our example, is a process of finding the best solution鈥攎aximum or minimum value鈥攇iven a specific set of constraints.
In practical applications, optimization helps to achieve efficient and cost-effective results in areas such as engineering, economics, and logistics.
The Lagrange multipliers method is particularly useful for constrained optimization problems because it helps find the extrema of a function subject to an equality constraint. Here's how it works in brief:
In practical applications, optimization helps to achieve efficient and cost-effective results in areas such as engineering, economics, and logistics.
The Lagrange multipliers method is particularly useful for constrained optimization problems because it helps find the extrema of a function subject to an equality constraint. Here's how it works in brief:
- Identify the function to maximize or minimize.
- Rewrite it using the constraint with the Lagrangian function, introducing the Lagrange multiplier \( \lambda \).
- Calculate gradients for each equation and set them equal, incorporating the multiplier \( \lambda \).
- Solve the resulting system of equations to find critical points which are then evaluated to find optimal values in terms of maximum or minimum.
Multivariable Calculus
Multivariable calculus extends calculus to functions of several variables. It allows for the mathematical modeling of more complex systems with multiple influencing factors.
Think of it as a branching out of single-variable calculus. Now, we deal with functions of two or more variables which are essential for problems in physics, engineering, and data science.
Multivariable calculus tools such as gradients, partial derivatives, and vector fields become powerful when combined with optimization techniques.
For example, in optimization, we often use gradients, which in this exercise are \( abla f(x, y) = (y, x) \) and \( abla g(x, y) = (10x, 2y) \). They point in the direction of greatest change and help identify the critical points by applying them in the method of Lagrange multipliers.
This framework is efficient for dealing with real-world scenarios where several variables must be considered simultaneously to optimize performance, costs, or other desired metrics.
Think of it as a branching out of single-variable calculus. Now, we deal with functions of two or more variables which are essential for problems in physics, engineering, and data science.
Multivariable calculus tools such as gradients, partial derivatives, and vector fields become powerful when combined with optimization techniques.
For example, in optimization, we often use gradients, which in this exercise are \( abla f(x, y) = (y, x) \) and \( abla g(x, y) = (10x, 2y) \). They point in the direction of greatest change and help identify the critical points by applying them in the method of Lagrange multipliers.
This framework is efficient for dealing with real-world scenarios where several variables must be considered simultaneously to optimize performance, costs, or other desired metrics.
