Problem 13 - Calculus with Concepts in Calculus

Question

Find the extreme values of \(f\) in the region described by the given inequalities. In each case assume that the extreme values exist. $$ f(x, y)=x y ; 2 x^{2}+y^{2} \leq 4 $$

Step-by-Step Solution

Answer

The extreme values of \(f(x, y) = xy\) are 0 and \(\frac{8}{5}\).

Step 1: Analyze the constraint

Given the constraint \(2x^2 + y^2 \leq 4\), this describes an ellipse centered at the origin. The region of interest is the area inside and including this ellipse.

Step 2: Find Critical Points

To find the critical points inside the region, we take the partial derivatives of \(f(x, y) = xy\). Compute \(f_x = y\) and \(f_y = x\). Set these equal to zero: \(y=0\) and \(x=0\). This gives a critical point at \((0,0)\).

Step 3: Evaluate the function at the critical point

Substitute the critical point \((0, 0)\) into the function: \(f(0, 0) = 0 \times 0 = 0\).

Step 4: Use the method of Lagrange Multipliers

To ensure that no other extrema exist on the boundary, introduce a Lagrange multiplier \(\lambda\) and set up the equations: \(f_x = \lambda g_x\) and \(f_y = \lambda g_y\) where \(g(x,y) = 2x^2 + y^2 - 4\). This gives \(y = 4\lambda x\) and \(x = 2\lambda y\).

Step 5: Solve the system with constraints

From \(y = 4\lambda x\) and \(x = 2\lambda y\), equate \(y/x = 4\lambda\) and \(x/y = 2\lambda\). Solving these gives \(\lambda = 0\) or \(x^2 = 2y^2\). Thus, substitute into the constraint: for \(\lambda = 0\), use values consistent with the boundary, and for \(x^2 = 2y^2\), substitute into \(2x^2 + y^2 = 4\).

Step 6: Evaluate boundary solutions

For \(\lambda = 0\), check solutions around the boundary like \(x = 2, y = 0\) and \(x= -2, y = 0\) gives \(f(2, 0) = 0\). For \(x^2 = 2y^2\), set \(x = \pm \sqrt{2}y\) in the constraint. Then, solve for \(y\) using \(2(\sqrt{2}y)^2 + y^2 = 4\) which simplifies to potential \(y\) values solving \(y(5) = 4\), yielding \(y = \pm \sqrt{\frac{4}{5}}\), \(x = \pm \sqrt{\frac{8}{5}}\). Check function \(f(x, y) = xy\) at these points. \(f(\sqrt{\frac{8}{5}}, \sqrt{\frac{4}{5}})\) and similar for negative counterparts.

Step 7: Identify extrema values

Evaluating all points provides maximum and minimum values of \(xy\). The evaluations give the values for maximum and minimum being generated from the points satisfying both the function maximal feasibilities and the boundary constraints.

Key Concepts

Extreme Values

In calculus, finding extreme values is a fundamental task in optimization problems. Extreme values, also known as extrema, are the highest or lowest points the function takes in a given region. These values can be classified into two types: maximum and minimum.
To find extreme values of a function, we often examine critical points. These critical points are where the first derivative (or in multivariable functions, partial derivatives) of the function equals zero. Once these points are found, we evaluate the function at these locations to identify potential extrema.
Additionally, when dealing with constraints, we need to consider points on the boundary of the region described by the constraints, as extrema can occur there too. In exercises, like the one involving the function \(f(x,y) = xy\) under the constraint \(2x^2 + y^2 \leq 4\), it is essential to analyze both the interior of the region and its boundary to find all possible extreme values.

Lagrange Multipliers

The method of Lagrange multipliers is a powerful technique used to locate extreme values of a function subject to equality constraints. It allows us to handle situations where we cannot simply use standard methods of finding critical points.

• This method involves introducing a new variable called the Lagrange multiplier (\(\lambda\)).
• For a function \(f(x,y)\) with a constraint \(g(x,y)=0\), we form the Lagrange function \(\mathcal{L}(x,y,\lambda) = f(x,y) - \lambda(g(x,y))\).
• We then find the partial derivatives of \(\mathcal{L}\) with respect to each variable and set them to zero to solve the system of equations.

By finding \(\lambda\) and the associated \(x\) and \(y\) values, we ensure any extrema are explored at both interior points and along the boundaries defined by the constraints. This approach was crucial in the exercise for determining potential extrema on the elliptical boundary as well.

Partial Derivatives

Partial derivatives represent how a multivariable function changes as one variable changes while all others are held constant. They are essential when dealing with functions with more than one variable, as they can specify the rate of change in different directions.
To find the partial derivatives of a function like \(f(x, y) = xy\), we take the derivative of \(f\) with respect to each variable, treating the other variable as a constant. So, the partial derivative with respect to \(x\) is \(f_x = y\), and with respect to \(y\) is \(f_y = x\).
Critical points, where partial derivatives equal zero, are candidates for extreme values. This process helps locate potential maximum or minimum points within the region of interest. By combining the results from partial derivatives with other methods, like Lagrange multipliers, we can thoroughly investigate the optimization problem.

Elliptical Constraints

Elliptical constraints are specific conditions that restrict the region where a function is evaluated. They are typically represented by equations of the form \(ax^2 + by^2 \leq c\), which describe an ellipse on a two-dimensional plane.
This type of constraint defines the boundary of the region, and it is crucial to consider it when finding extrema since they can occur at any point along this boundary as well as within the region.
In our original exercise, the constraint \(2x^2 + y^2 \leq 4\) outlines an elliptical shape centered at the origin. This restricts our function \(f(x,y) = xy\) to the interior and surface of this ellipse. Therefore, any extreme value needs to satisfy both the functional and the constraint conditions.
By properly considering the elliptical boundary and applying techniques like inspecting critical points and using Lagrange multipliers, one can identify all possible extreme values effectively. Optimization problems often depend heavily on such constraints to define feasible regions and ensure realistic solutions.