Question

Find the absolute extrema of the function over the region \(R .\) (In each case, \(R\) contains the boundaries.) Use a computer algebra system to confirm your results. \(f(x, y)=2 x-2 x y+y^{2}\) \(R\) : The region in the \(x y\) -plane bounded by the graphs of \(y=x^{2}\) and \(y=1\)

Short answer

The absolute maximum of the given function on region \(R\) is \(5\) at point \((-1, 1)\) and the absolute minimum is \(0\) at points \((0, 0)\) and \((1, 1)\).

Step-by-step solution

Find the critical points

The critical points occur where the gradient of the function is zero or undefined. To find these points, we first derive the function \(f(x, y)=2 x-2 x y+y^{2}\) with respect to \(x\) and \(y\) to get \(f_x(x, y)=2-2y\) and \(f_y(x, y)=-2x+2y\). Then, we set these equations to zero to find the \(x\) and \(y\) values of the critical points, which yields: \(2-2y=0\) and \(-2x+2y=0\). Hence, solving these equations gives us one critical point \((x,y)=(1,1)\).

Perform the second derivative test

The second derivative test uses the second partial derivatives of the function to determine the type of a critical point. First, find the second partial derivatives \(f_{xx} = 0, f_{yy} = 2, f_{xy} = f_{yx} = -2\). Then calculate the determinant of the Hessian matrix (denoted by \(D\)) which is \(f_{xx}*f_{yy} - (f_{yx}*f_{xy}) = 0*2 - ((-2)(-2)) = -4\). Since \(D < 0\), the critical point \((1,1)\) is a saddle point and thus neither a local maximum nor a minimum.

Evaluate the function on the boundaries

The region \(R\) is bounded by the graphs of \(y=x^{2}\) and \(y=1\). Substituting these into the function will give the values of the function on the boundaries. For the boundary \(y=x^{2}\), the function \(f(x, y) = 2x - 2x*x^{2}+x^{4} = x^{4}-2x^3+2x\). The critical points can be found by setting the derivative to zero, yielding the solutions \(x=0, 1, \text{and } -1\). For \(y = 1\), the function becomes \(f(x, y) = 2x - 2x*1 +1 = 2x-2x+1 = 1\), which is a constant function. Hence, the function values on the boundaries of \(R\) are \(f(0,0)=0, f(1,1)=1, f(-1,1)=5\)

Identify the absolute extrema

Compare the function values at the critical point and on the boundaries to determine the absolute extrema. The maximum function value is \(5\) at point \((-1, 1)\), and the minimum is \(0\) at points \((0, 0)\) and \((1, 1)\). Hence on \(R\), the absolute maximum of \(f\) is \(5\) at \((-1,1)\) and the absolute minimum is \(0\) at \((0,0)\) and \((1,1)\).

Key concepts

Critical Points

Understanding critical points is foundational in calculus for identifying potential local maxima and minima of a function. To find critical points of a function of two variables, such as f(x, y), we look for points where the gradient vector (consisting of all partial derivatives) is equal to zero or the partial derivatives do not exist.

For a function f(x, y), we calculate the first partial derivatives, termed f_x and f_y, with respect to 'x' and 'y' respectively. The locations where both partial derivatives equal zero simultaneously are the critical points. In our example with the function f(x, y) = 2x - 2xy + y², setting f_x and f_y equal to zero gave us the critical point at (1, 1), a potential location for extremum within the bordered region R.

Second Derivative Test

Once critical points are discovered, the second derivative test is a valuable tool to classify them. This test is based on the second partial derivatives of a function f(x, y). For a smooth function, we calculate the second partial derivatives f_xx, f_yy, and the mixed partial derivatives f_xy (f_yx by Clairaut's theorem).

In our function, the second partial derivatives are f_xx = 0, f_yy = 2, and f_xy = f_yx = -2. The second derivative test then uses these to calculate the determinant of the Hessian matrix. If the determinant is positive, and f_xx is positive, the critical point is a local minimum; if f_xx is negative, it's a local maximum. If the determinant is negative, the point is a saddle point, which is the case at point (1, 1) in our example, meaning it is not an extremum.

Partial Derivatives

The concept of partial derivatives represents the rate of change of a multivariable function with respect to one variable at a time. While a regular derivative applies to functions with a single variable, partial derivatives extend this concept to functions of multiple variables, like our function f(x, y).

To calculate a partial derivative, you treat all other variables as constants and differentiate with respect to the variable of interest. For instance, the partial derivative of f(x, y) with respect to 'x' is denoted as f_x and shows how f changes as 'x' changes while keeping 'y' constant. Similarly, f_y expresses the rate of change of f with respect to 'y'. Knowing how to calculate and interpret partial derivatives is vital in finding potential extrema of multivariable functions.

Hessian Matrix

The Hessian matrix is a square matrix composed of second-order partial derivatives of a scalar-valued function, and it provides profound insight in multivariable calculus, especially in optimization problems. It helps to determine the curvature at a point on a surface described by a multivariable function.

For a function with two variables f(x, y), the Hessian matrix is a 2x2 matrix that contains all the second partial derivatives: f_xx, f_xy, f_yx, and f_yy. In mathematical notation, the Hessian is represented as:
\[ H(f(x, y)) = \begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \end{bmatrix} \]
The determinant of this matrix, evaluated at a critical point, is used in the second derivative test to classify the critical point. A positive determinant suggests a local minimum or maximum, while a negative determinant denotes a saddle point. It is an essential concept for verifying the nature of the critical points found in the previous steps.