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. $$ \begin{array}{l} f(x, y)=x^{2}-4 x y+5 \\ R=\\{(x, y): 0 \leq x \leq 4,0 \leq y \leq \sqrt{x}\\} \end{array} $$

Short answer

The absolute minimum of the function \(f(x,y) = x^2 - 4xy + 5\) over the region R is 5, attained at the point (0,0). The absolute maximum over the region R is 9, attained at the point (4,2)

Step-by-step solution

Identify and Evaluate the Critical Points

The critical points of the function are found by solving the system of equations given by the first derivatives of the function: \(f_x = 0\) and \(f_y = 0\) . For the function \(f(x,y) = x^2 - 4xy + 5\), we have \(f'_x=2x-4y\) and \(f'_y=-4x\). Setting these equal to zero gives \(2x-4y=0\) and \(-4x=0\). Solving this system of equations we find only one critical point (0,0). We evaluate the function at this point to find \(f(0,0)=5\).

Evaluate the Function on the Boundary

The boundary of the region R is defined by the equations \(x=0\), \(x=4\), and \(y=\sqrt{x}\). We evaluate the function on this boundary by substituting these into the function and simplifying. For \(x=0\), we get \(f(0,y)=5\); for \(x=4\), we get \(f(4,y)=4y^2 - 16y +5\), and for \(y=\sqrt{x}\), we get \(f(x,\sqrt{x})=x^2 - 4x\sqrt{x} +5\).

Find the Absolute Extrema

The absolute extrema of the function on the region R are the largest and smallest values the function takes on the critical points and the boundary. Checking the values we've found, we see that on the boundary, the function is always equal to 5 or more greater. Hence, the minimum value of the function in the region R is \(f(0,0)=5\), attained at the point (0,0). The maximum value would be found at \(f(4,\sqrt{4}) = f(4,2) = 9\).

Key concepts

Critical Points in Calculus

In calculus, a critical point of a continuous function is a point at which the derivative is zero or undefined. To find the critical points of a function, you first determine its derivative and then solve for the points where this derivative equals zero.

For a function of one variable, like \(f(x)\), the critical points are found by solving \(f'(x) = 0\). In the context of functions of multiple variables, such as \(f(x, y)\), you need to look at the partial derivatives with respect to each variable. In our case, the partial derivatives of \(f(x, y) = x^2 - 4xy + 5\) are \(f'_x = 2x - 4y\) and \(f'_y = -4x\). Setting these partial derivatives to zero gives us the system of equations to solve for the critical points. Here, we found the only critical point to be (0,0), by solving \(2x - 4y = 0\) and \(-4x = 0\). It's essential to evaluate the function at each critical point to determine if they correspond to a minimum, maximum, or neither.

Evaluating our function at the critical point (0,0) yields \(f(0,0) = 5\), which is an important step in determining the absolute extrema of the function.

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. They are used to measure the rate of change of a multivariable function with respect to one variable while keeping other variables constant.

The general form of a partial derivative of a function \( f(x, y) \) with respect to \( x \) is denoted as \( \frac{\partial f}{\partial x} \) or \( f'_x \) as a shorthand notation. To find it, you differentiate \( f \) with respect to \( x \), treating other variables as constants. Similarly, \( \frac{\partial f}{\partial y} \) or \( f'_y \) represents the partial derivative with respect to \( y \).

In our example, we determined the partial derivatives \(f'_x = 2x - 4y\) and \(f'_y = -4x\). These derivatives are critical to finding where the gradient of the function \(f\) is zero, leading us to potential extrema which are points of interest for optimization.

Optimization in Multivariable Calculus

Optimization in multivariable calculus involves finding the maximum or minimum values of functions with several variables. This process is crucial in many applications, from physics to economics, where determining the best outcome under given conditions is desired.

After identifying the critical points, the next step involves checking the boundaries of the region under consideration. This is because absolute extrema can occur at critical points or along the boundaries of the domain. For the function \(f(x, y) = x^2 - 4xy + 5\) over the region \(R\), defined by \(0 \leq x \leq 4\) and \(0 \leq y \leq \sqrt{x}\), we examine the boundary by inserting these constraints into the function and simplifying.

Once both critical points and boundary values are assessed, you compare them to determine the absolute maximum and minimum, known as the absolute extrema. In our exercise, the minimum value of the function within the region is at the critical point (0,0) where \(f(0,0) = 5\), and the maximum is found on the boundary at the point (4,2), where \(f(4,2) = 9\). This illustrates how optimization in multivariable calculus combines the use of partial derivatives and boundary evaluations to solve for the most extreme values a function can take.