Question

Use Lagrange multipliers in the following problems. When the constraint curve is unbounded, explain why you have found an absolute maximum or minimum value. Minimum distance to a cone Find the points on the cone \(z^{2}=x^{2}+y^{2}\) closest to the point (1,2,0)

Short answer

Question: Find the point on the cone \(z^2 = x^2 + y^2\) that is closest to the point (1, 2, 0). Answer: The point on the cone closest to (1, 2, 0) is \((-\frac{2}{3}, \frac{1}{3}, \frac{1}{3})\).

Step-by-step solution

Define the Objective Function and Constraint

Let point P(x, y, z) be a point on the cone \(z^2 = x^2 + y^2\). We want to minimize the distance between this point and the given point (1, 2, 0). The distance between two points in 3D space is given by the following formula: Distance: \(D = \sqrt{(x - 1)^2 + (y - 2)^2 + z^2}\) Since square root is a monotonic function, want to minimize the square of the distance in our objective function. Objective function (D^2): \(f(x, y, z) = (x - 1)^2 + (y - 2)^2 + z^2\) The constraint equation is given by the equation of the cone: Constraint: \(g(x, y, z) = z^2 - x^2 - y^2\)

Set Up the Lagrange Multipliers

Now that we have the objective function and the constraint, we can set up the Lagrange equation: \(L(x, y, z, \lambda) = f(x, y, z) - \lambda \cdot g(x, y, z)\) Substitute f and g from Step 1: \(L(x, y, z, \lambda) = (x - 1)^2 + (y - 2)^2 + z^2 - \lambda(z^2 - x^2 - y^2)\)

Compute the Gradient and Set Equations

To find the critical points using the Lagrange multiplier, we have to compute the gradient of L(x, y, z, λ) and set each component to 0. \(\nabla L(x, y, z, \lambda) = \left(\frac{\partial L}{\partial x}, \frac{\partial L}{\partial y}, \frac{\partial L}{\partial z}, \frac{\partial L}{\partial \lambda} \right) = (0, 0, 0, 0)\) Compute each partial derivative: \(\frac{\partial L}{\partial x} = 2(x - 1) + 2\lambda x\) \(\frac{\partial L}{\partial y} = 2(y - 2) + 2\lambda y\) \(\frac{\partial L}{\partial z} = 2z - 2\lambda z\) \(\frac{\partial L}{\partial \lambda} = z^2 - x^2 - y^2\) Set each derivative equal to zero and obtain the following equations: 1) \(2(x - 1) + 2\lambda x = 0\) 2) \(2(y - 2) + 2\lambda y = 0\) 3) \(2z - 2\lambda z = 0\) 4) \(z^2 - x^2 - y^2 = 0\)

Solve the Equations

Now we need to solve the system of equations obtained in Step 3. First, let's solve equation 3 for λ: 3) \(\lambda = \frac{z}{z-1}\) Substitute the value of λ into the first and second equations: 1) \(2(x - 1) + 2\frac{z}{z-1}x = 0\) 2) \(2(y - 2) + 2\frac{z}{z-1} y = 0\) Now solve equations 1 and 2 for x and y, respectively, in terms of z: \(x = \frac{(z-1)}{z} = 1 - \frac{1}{z}\) \(y = \frac{2(z-1)}{z} = 2 - \frac{2}{z}\) Substitute x and y expressions into equation 4: \((2 - \frac{2}{z})^2 - (1 - \frac{1}{z})^2 - z^2 = 0\) Solving this equation yields the solutions z = −1 and z = 1/3. Note that z = -1 corresponds to a point below the cone's vertex and doesn't lie on the surface of the cone, so we are not considering this solution. The only valid solution is z = 1/3. Now substitute this value into the expressions for x and y: \(x = 1 - \frac{1}{1/3} = -\frac{2}{3}\) \(y = 2 - \frac{2}{1/3} = \frac{1}{3}\)

Verify the Solution

We've found the point \((-\frac{2}{3}, \frac{1}{3}, \frac{1}{3})\) as the closest point to (1, 2, 0) on the cone \(z^2 = x^2 + y^2\). To verify that we have found an absolute minimum value, we can examine the nature of the cone and the problem statement. It turns out that the constraint curve (the cone) is unbounded, and the distance function D is always positive. Therefore, we have found an absolute minimum.

Key concepts

Minimize Distance

In many real-world scenarios, including geometric problems and operations research, the goal is often to find the shortest path or minimum distance between points. When the points lie on a surface or curve defined by a mathematical constraint, this becomes a more complex problem of optimization under constraint.

For instance, when we aim to minimize the distance between a given point and a cone's surface, we essentially search for the point on the cone that is nearest to the specified location. This process involves creating an objective function that represents the square of the distance between the two points, since minimizing the square also minimizes the actual distance. Then we look for its minimum value while taking into account the constraint that the point must lie on the cone. This leads us into the realm of calculus and, more specifically, the method of Lagrange multipliers.

Constraint Optimization

Constraint optimization problems are at the heart of many fields such as economics, engineering, and mathematics. They involve finding the optimal value (either maximum or minimum) of an objective function within a set of constraints. In calculus, the method of Lagrange multipliers is a powerful tool for solving such problems.

The beauty of Lagrange multipliers lies in its introduction of an auxiliary variable, the 'multiplier', which incorporates the constraint into the optimization problem. This connective tissue between the objective function and the constraint allows us to solve for the points where the objective function may have optimal values. Deep understanding of constraint optimization is vital for students not only in academic problems but also in real-life applications where resources or conditions are limited.

Calculus

Calculus is the branch of mathematics that studies how things change. It provides a framework for modeling systems in which optimization problems can be defined and solved. These solutions can be simple when dealing with linear equations or systems without constraints, but most often in calculus, real-world problems require techniques for managing more complex situations.

Lagrange multipliers, for instance, is a technique that stands out in calculus for handling the difficulty of constraint optimization problems. It requires the calculation of partial derivatives and critical points of the Lagrangian, which is a function combining the original objective function with the constraints. Successfully applying calculus concepts like derivatives and understanding the meaning of critical points are crucial for tackling these challenges.

3D Space Distance Calculation

When we calculate distances in 3D space, we extend the familiar Pythagorean theorem into three dimensions. The distance formula \(D = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2}\) serves as the objective function for many optimization tasks, including our example of finding the shortest distance to a cone from a point.

To handle such problems, a clear understanding of 3D space and its geometric interpretations helps visualize the circumstances under which we are optimizing. One must accurately represent both the objective function and the constraints in three-dimensional terms to navigate toward the solution. The objective function encompasses the essence of our problem, which, coupled with calculus tools, paves the way towards finding a minimum or maximum value under given constraints.