Problem 13 - Active Calculus Multivariable

Question

Use the method of Lagrange multipliers to find the point on the line \(x-2 y=5\) that is closest to the point \((1,3) .\) To do so, respond to the following prompts. a. Write the function \(f=f(x, y)\) that measures the square of the distance from \((x, y)\) to (1,3) . (The extrema of this function are the same as the extrema of the distance function, but \(f(x, y)\) is simpler to work with.) b. What is the constraint \(g(x, y)=c ?\) c. Write the equations resulting from \(\nabla f=\lambda \nabla g\) and the constraint. Find all the points \((x, y)\) satisfying these equations. d. Test all the points you found to determine the extrema.

Step-by-Step Solution

Answer

The point on the line \(x-2y=5\) that is closest to the point \((1,3)\) using the method of Lagrange multipliers is \((3, -1)\).

Step 1: Define the function f(x, y)#

Given that \(f(x, y)\) is the square of the distance between a point \((x, y)\) and \((1, 3)\), we can use the Euclidean distance formula and square it: \(f(x, y) = (x-1)^2 + (y-3)^2\) #b. Determine the constraint g(x, y)=c#

Step 2: Find the constraint g(x, y)=c#

The constraint is given by the equation of the line: \(x-2y=5\). So, we can define: \(g(x, y) = x - 2y\) And the constraint is: \(g(x, y) = c\) Where \(c = 5\). #c. Write and solve the Lagrange multiplier equations#

Step 3: Write the Lagrange multiplier equations#

Using the method of Lagrange multipliers, we have: \(\nabla f(x, y) = \lambda \nabla g(x, y)\) First, we need to calculate the gradients for both f and g functions: \(\nabla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right) = (2(x-1), 2(y-3))\) \(\nabla g(x, y) = \left(\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}\right) = (1, -2)\) Now, we can write the equations: \(2(x-1) = \lambda\) \(2(y-3) = -2\lambda\) And don't forget the constraint: \(x - 2y = 5\)

Step 4: Solve the Lagrange multiplier equations#

Now, we will solve the system of equations: \(2(x-1) = \lambda\) \(2(y-3) = -2\lambda\) \(x - 2y = 5\) First, we can solve the constraint equation for x and substitute it in the first Lagrange multiplier equation: \(x = 2y + 5\) \(2(2y+5-1) = \lambda \) \(4y+8 = \lambda\) Now, we can substitute this into the second Lagrange multiplier equation: \(2(y-3) = -2 (4y+8)\) \(2y - 6 = -8y - 16\) Add 8y to both sides and add 6 to both sides: \(10y = -10\) Now, divide by 10: \(y = -1\) Now, we can use the constraint equation with our y value to find the x value: \(x=2(-1)+5\) \(x=3\) So, we have found that the candidate point is \((3,-1)\). #d. Testing the candidate points#

Step 5: Determine the extrema#

As there is only one candidate point \((3,-1)\), we can be sure that this is the point on the line that is closest to the point \((1,3)\), without the need to test it for an extreme. We can visualize the solution on the Cartesian plane, and it's evident that this is the point where the line is closest to the given point. So, the point on the line \(x-2y=5\) that is closest to the point \((1,3)\) is \((3, -1)\).

Key Concepts

Euclidean distance

To better understand the process of finding the closest point on a line, let's explore the concept of Euclidean distance. Euclidean distance measures the straight-line distance between two points in multidimensional space. In our two-dimensional case, the distance between points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]However, to simplify calculations, especially in optimization problems, we often use the square of this distance. It removes the square root, making the function smoother to work with:\[ f(x, y) = (x - x_1)^2 + (y - y_1)^2 \]In this problem, the function \( f(x, y) = (x - 1)^2 + (y - 3)^2 \) is used, where \( (1,3) \) is the point we are comparing distances from.

Gradient vectors

Understanding gradient vectors is key to solving optimization problems using Lagrange multipliers. A gradient vector provides the direction in which a function increases most rapidly. For a function \( f(x,y) \), its gradient, denoted as \( abla f(x, y) \), is a vector of partial derivatives:\[ abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \]This gradient points towards the steepest ascent of the function. In Lagrange multiplier problems, we find that the gradient of the objective function \( abla f(x, y) \) is proportional to the gradient of the constraint function \( abla g(x, y) \), represented as:\[ abla f = \lambda abla g \]In this exercise, the gradients become:

• \( abla f = (2(x-1), 2(y-3)) \)
• \( abla g = (1, -2) \)

This forms the basis for creating a system of equations to find critical points.

Extreme value problems

Extreme value problems involve finding maximum or minimum values of a function subject to certain constraints. In this context, we use the method of Lagrange multipliers. It allows us to solve these problems by introducing a multiplier that helps manage constraints while seeking extrema.In our exercise, we aim to find a point on the line \( x - 2y = 5 \) that is closest to \( (1,3) \). Instead of directly finding minima of the distance, we solve an extreme value problem for the squared distance function. This bypasses the complication of square roots and maintains the same extrema attributes.The step involves forming a system of equations derived from setting the gradient of the squared distance function equal to a multiple of the gradient of the constraint function; essentially:\[ abla f = \lambda abla g \]By solving this along with the constraint equation, we arrive at possible extrema points.

System of equations

A system of equations is a set of equations with shared variables, which must be solved together to find common solutions. In the context of Lagrange multipliers, we derive a system of equations from the gradients and constraints. For our example:

• \( 2(x-1) = \lambda \)
• \( 2(y-3) = -2\lambda \)
• Constraint: \( x - 2y = 5 \)

Solving this system requires substitution and elimination strategies. First, we express \( \lambda \) in terms of \( x \) from the first equation and substitute it into the second. Simultaneously, solve the constraint for one variable to reduce complexity. Integration of these processes yields the solution \( (x, y) = (3, -1) \), confirming that it is the extrema point where the line is closest to \( (1,3) \). This systematic approach illustrates the powerful insight that Lagrange multipliers and systems of equations provide in optimization challenges.