Problem 385 - Calculus Volume 3

Question

Use the method of Lagrange multipliers to solve the following applied problems. Find the minimum distance from point (0,1) to the parabola \(x^{2}=4 y\).

Step-by-Step Solution

Answer

The minimum distance is 1.

Step 1: Define the Objective Function

We are tasked with finding the minimum distance from the point \((0, 1)\) to the parabola defined by the equation \(x^2 = 4y\). To start, we need an expression for the distance between any point \((x, y)\) on the parabola and the point \((0, 1)\). The distance formula is given by \(d = \sqrt{(x - 0)^2 + (y - 1)^2}\). To simplify calculations using the Lagrange multipliers, we minimize the square of the distance function, \(f(x, y) = x^2 + (y - 1)^2\).

Step 2: Set Up the Constraint

The constraint is given by the equation of the parabola \(g(x, y) = x^2 - 4y = 0\). This constraint describes the path where the point \((x, y)\) must lie.

Step 3: Compute the Gradients

The method of Lagrange multipliers involves setting the gradient of the objective function equal to a scalar multiple (\(\lambda\)) of the gradient of the constraint. Compute the gradients: \(abla f = (2x, 2(y - 1))\) and \(abla g = (2x, -4)\).

Step 4: Set Up the Lagrange Equations

Equate the gradients: \((2x, 2(y - 1)) = \lambda(2x, -4)\).This results in two equations: \(2x = \lambda 2x\) and \(2(y - 1) = \lambda (-4)\). Also, consider the constraint \(x^2 = 4y\).

Step 5: Analyze and Solve the Equations

From \(2x = \lambda 2x\), if \(x eq 0\), \(\lambda = 1\). Continuing, it leads to \(2(y - 1) = -4\), which simplifies to \(y = -1\). Substitute \(y = -1\) into the constraint \(x^2 = 4(-1)\), giving us \(x^2 = -4\), which is impossible for real numbers. Thus, our assumption of \(x eq 0\) fails, implying \(x = 0\). Substitute \(x = 0\) in the constraint \(x^2 = 4y\) gives \(0 = 4y\) or \(y = 0\).

Step 6: Calculate the Distance

The only feasible solution is \((x, y) = (0, 0)\). Compute the distance from \((0, 0)\) to \((0, 1)\) using the distance formula: \(d = \sqrt{(0 - 0)^2 + (0 - 1)^2} = \sqrt{1} = 1\).

Key Concepts

Constraint Optimization

To solve problems involving optimizing a function subject to certain constraints, we often use the method of Lagrange multipliers. This powerful technique is particularly useful when dealing with curves or surfaces described by algebraic equations. In our exercise, we're dealing with a point located at

• (0,1)

and a parabola defined by

• \(x^2 = 4y\)

This is a classic example of constraint optimization, where the distance function we want to minimize is subject to the point lying on the curve specified by the parabola.The primary idea is to convert the constrained problem into a form that is easier to solve. This involves introducing an auxiliary variable, \(\lambda\), which acts as a weighting factor to balance the contributions of the objective function and the constraint.
The constraint itself, represented here as \(g(x, y) = x^2 - 4y\), ensures that the solution obeys the

• given path of the parabola.

The key to this optimization is not altering the nature of the original constraint but working around it strategically using mathematical tools and

• concepts like gradients and scalar multiplication.

Distance Minimization

Finding the shortest path or minimum distance between a point and a curve is a frequent problem in optimization. In our exercise, we need to determine the closest distance from the point

• (0,1) to the parabola \(x^2 = 4y\).

Here, instead of minimizing the actual distance, which involves a square root, we minimize its square to simplify calculations. This is common since the square of a non-negative number is also minimized where the original number is minimized. Thus, we consider the function

• \(f(x, y) = x^2 + (y - 1)^2\)

as our objective. Our goal becomes minimizing \(f\) subject to the constraint \(x^2 = 4y\).This method balances calculation simplicity with accurate results. Remember that the absolute minimum distance is realized when the square of the distance is minimized to zero, or otherwise, the smallest possible positive value.

Gradient Calculation

Gradients play a pivotal role in the method of Lagrange multipliers. They point in the direction of the greatest rate of increase of a function. In the context of our exercise, we examine the gradients of both the objective function and the constraint:

• The gradient of the objective function, \(abla f = (2x, 2(y - 1))\), represents how \(f\) changes with small variations in \(x\) and \(y\).
• The gradient of the constraint function, \(abla g = (2x, -4)\), indicates how the constraint \(x^2 = 4y\) changes.

By setting \(abla f = \lambda abla g\), we create a system of equations that describes the behavior of the original constrained optimization problem. This equation essentially states that, at the optimum point, the rate of change in the objective function along any direction that maintains the constraint should not exceed a certain threshold.Through careful analysis and solution of these equations, one can find the value of \(\lambda\) and, subsequently, the values of \(x\) and \(y\) that satisfy both the constraint and provide extremum for the objective function. In our case, after a thorough exploration, it was found that the true viable solution is the point \((0,0)\), correctly reflecting the constraints and minimizing the associated function.