(Source: www.youtube.com)

Contents

Examine critical points and boundary points to find absolute maximum and minimum values for a function of two variables. One of the most useful applications for derivatives of a function of one variable is the determination of maximum and/or minimum values.

This application is also important for functions of two or more variables, but as we have seen in earlier sections of this chapter, the introduction of more independent variables leads to more possible outcomes for the calculations. The main ideas of finding critical points and using derivative tests are still valid, but new wrinkles appear when assessing the results.

For functions of a single variable, we defined critical points as the values of the variable at which the function's derivative equals zero or does not exist. For functions of two or more variables, the concept is essentially the same, except for the fact that we are now working with partial derivatives.

Then, multiply each equation by its common denominator: We must also check for the possibility that the denominator of each partial derivative can equal zero, thus causing the partial derivative not to exist.

There are no points in \(\math{R}^2\) that make either partial derivative not exist. Figure \(\PageIndex{1}\) shows the behavior of the surface at the critical point.

(Source: www.youtube.com)

Hint Calculate \(f_x(x, y)\) and \(f_y(x, y)\), then set them equal to zero. The main purpose for determining critical points is to locate relative maxima and minima, as in single-variable calculus.

The number \(f(x_0,y_0)\) is called a local minimum value. If the preceding inequality holds for every point \((x, y)\) in the domain of \(f\), then \(f\) has a global minimum (also called an absolute minimum) at \((x_0,y_0)\).

It attains its minimum value at the boundary of its domain, which is the circle \(x^2+y^2=16.\)In Calculus 1, we showed that extreme of functions of one variable occur at critical points. The same is true for functions of more than one variable, as stated in the following theorem.

Let \(z=f(x, y)\) be a function of two variables that is defined and continuous on an open set containing the point \((x_0,y_0)\). Therefore, the existence of a critical value at \(x=x_0\) does not guarantee a local extreme at \(x=x_0\).

An example of a saddle point appears in the following figure. This is because the first partial derivatives of f\((x, y)=x^2y^2\) are both equal to zero at this point, but it is neither a maximum nor a minimum for the function.

(Source: www.youtube.com)

Furthermore, the vertical trace corresponding to \(y=0\) is \(z=x^2\) (a parabola opening upward), but the vertical trace corresponding to \(x=0\) is \(z=y^2\) (a parabola opening downward). The second derivative test for a function of one variable provides a method for determining whether an extreme occurs at a critical point of a function.

When extending this result to a function of two variables, an issue arises related to the fact that there are, in fact, four different second-order partial derivatives, although equality of mixed partials reduces this to three. The second derivative test for a function of two variables, stated in the following theorem, uses a discriminant \(D\)that replaces \(f''(x_0)\) in the second derivative test for a function of one variable.

Let \(z=f(x, y)\) be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point \((x_0,y_0)\). Figure \(\PageIndex{4}\) : The second derivative test can often determine whether a function of two variables has local minima (a), local maxima (b), or a saddle point (c). To apply the second derivative test, it is necessary that we first find the critical points of the function.

Let \(z=f(x, y)\) be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point \((x_0,y_0).\) To apply the second derivative test to find localextrema, use the following steps: Calculate the discriminant \(D=f_{xx}(x_0,y_0)f_{by}(x_0,y_0)\big(f_{by}(x_0,y_0)\big)^2\) for each critical point of \(f\).

Apply the four cases of the test to determine whether each critical point is a local maximum, local minimum, or saddle point, or whether the theorem is inconclusive. Step 1 of the problem-solving strategy involves finding the critical points of \(f\).

(Source: www.youtube.com)

Step 3 states to apply the four cases of the test to classify the function's behavior at this critical point. Therefore, \(f\) has a local minimum at \((1,2)\) as shown in the following figure.

\ g_{by}(x, y) &=2\\ g_{by}(x, y) &=2. In step 3, we note that, applying Note to point \(\left(1,\franc{5}{2}\right)\) leads to case \(3\), which means that \(\left(1,\franc{5}{2}\right)\) is a saddle point.

Use the second derivative test to find the localextrema of the function Hint Follow the problem-solving strategy for applying the second derivative test.

Answer \(\left(\franc{4}{3}, \franc{1}{3}\right)\) is a saddle point, \(\left(\franc{3}{2}, \franc{3}{8}\right)\) is a local maximum. When finding global extreme of functions of one variable on a closed interval, we start by checking the critical values over that interval and then evaluate the function at the endpoints of the interval.

Then, it is necessary to find the maximum and minimum value of the function on the boundary of the set. A continuous function \(f(x, y)\) on a closed and bounded set \(D\) in the plane attains an absolute maximum value at some point of \(D\) and an absolute minimum value at some point of \(D\).

(Source: www.youtube.com)

Assume \(z=f(x, y)\) is a differentiable function of two variables defined on a closed, bounded set \(D\). But an interior point \((x_0,y_0)\) of \(D\) that’s an absolute extreme is also a local extreme; hence, \((x_0,y_0)\) is a critical point of \(f\) by Fermat’s theorem.

Problem-Solving Strategy: Finding Absolute Maximum and Minimum Values Let \(z=f(x, y)\) be a continuous function of two variables defined on a closed, bounded set \(D\), and assume \(f\) is differentiable on \(D\).

To find the absolute maximum and minimum values of \(f\) on \(D\), do the following: Determine the maximum and minimum values of \(f\) on the boundary of its domain.

Finding the maximum and minimum values of \(f\) on the boundary of \(D\) can be challenging. If the boundary is a rectangle or set of straight lines, then it is possible to parameterize the line segments and determine the maxima on each of these segments, as seen in Example \(\PageIndex{3}\).

The same approach can be used for other shapes such as circles and ellipses. If the boundary of the set \(D\) is a more complicated curve defined by a function \(g(x, y)=c\) for some constant \(c\), and the first-order partial derivatives of \(g\) exist, then the method of Lagrange multipliers can prove useful for determining the extreme of \(f\) on the boundary which is introduced in Lagrange Multipliers.

(Source: www.bartleby.com)

Use the problem-solving strategy for finding absolute extreme of a function to determine the absolute extreme of each of the following functions: Using the problem-solving strategy, step \(1\) involves finding the critical points of \(f\) on its domain.

Setting them equal to zero yields the system of equations The next step involves finding the extreme of \(f\) on the boundary of its domain.

The boundary of its domain consists of four line segments as shown in the following graph: Figure \(\PageIndex{7}\): Graph of the domain of the function \(f(x, y)=x^22xy+4y^24x2y+24.\)\(L_1\) is the line segment connecting \((0,0)\) and \((4,0)\), and it can be parameterized by the equations \(x(t)=t, y(t)=0\) for \(0t4\).

Differentiating \(g\) leads to \(g(t)=2t4.\) Therefore, \(g\) has a critical value at \(t=2\), which corresponds to the point \((2,0)\). \(L_2\) is the line segment connecting \((4,0)\) and \((4,2)\), and it can be parameterized by the equations \(x(t)=4,y(t)=t\) for \(0t2.\) Again, define \(g(t)=f\big(x(t), y(t)\big).\) This gives \(g(t)=4t^210t+24.\) Then, \(g(t)=8t10\).

G has a critical value at \(t=\franc{5}{4}\), which corresponds to the point \(\left(0,\franc{5}{4}\right).\) Calculating \(f\left(0,\franc{5}{4}\right)\) gives the \(z\)-value \(27.75\). \(L_3\) is the line segment connecting \((0,2)\) and \((4,2)\), and it can be parameterized by the equations \(x(t)=t, y(t)=2\) for \(0t4.\) Again, define \(g(t)=f\big(x(t), y(t)\big).\) This gives \(g(t)=t^28t+36.\) The critical value corresponds to the point \((4,2).\) So, calculating \(f(4,2)\) gives the \(z\)-value \(20\).

(Source: www.youtube.com)

\(L_4\) is the line segment connecting \((0,0)\) and \((0,2)\), and it can be parameterized by the equations \(x(t)=0,y(t)=t\) for \(0t2.\) This time, \(g(t)=4t^22t+24\) and the critical value \(t=\franc{1}{4}\) correspond to the point \(\left(0,\franc{1}{4}\right)\). The absolute maximum value is \(36\), which occurs at \((0,2)\), and the global minimum value is \(20\), which occurs at both \((4,2)\) and \((2,0)\) as shown in the following figure.

The next step involves finding the extreme of g on the boundary of its domain. The boundary of its domain consists of a circle of radius \(4\) centered at the origin as shown in the following graph.

The absolute maximum of \(g\) is approximately equal to 44.844, which is attained at the boundary point \(\left(\franc{8\sort{13}}{13}, \franc{12\sort{13}}{13}\right)\). These are the absolute extreme of \(g\) on \(D\) as shown in the following figure.

Use the problem-solving strategy for finding absolute extreme of a function to find the absolute extreme of the function Hint Calculate \(f_x(x, y)\) and \(f_y(x, y)\), and set them equal to zero.

The maximum number of golf balls that can be produced and sold is \(50,000\), and the maximum number of hours of advertising that can be purchased is \(25\). Figure \(\PageIndex{11}\): (credit: modification of work by oatsy40, Flickr)Using the problem-solving strategy, step \(1\) involves finding the critical points of \(f\) on its domain.

(Source: www.youtube.com)

Setting them equal to zero yields the system of equations The domain of this function is \(0x50\) and \(0y25\) as shown in the following graph.

Figure \(\PageIndex{12}\): Graph of the domain of the function \(f(x, y)=48x+96yx^22xy9y^2.\)\(L_1\) is the line segment connecting \((0,0)\) and \((50,0), \) and it can be parameterized by the equations \(x(t)=t, y(t)=0\) for \(0t50.\) We then define \(g(t)=f(x(t), y(t)):\) \(L_2\) is the line segment connecting \((50,0)\) and \((50,25)\), and it can be parameterized by the equations \(x(t)=50,y(t)=t\) for \(0t25\).

\(L_3\) is the line segment connecting \((0,25)\) and \((50,25)\), and it can be parameterized by the equations \(x(t)=t, y(t)=25\) for \(0t50\). \(L_4\) is the line segment connecting \((0,0)\) to \((0,25)\), and it can be parameterized by the equations \(x(t)=0,y(t)=t\) for \(0t25\).

Therefore, a maximum profit of \($648,000\) is realized when \(21,000\) golf balls are sold and \(3\) hours of advertising are purchased per month as shown in the following figure. To find extreme of functions of two variables, first find the critical points, then calculate the discriminant and apply the second derivative test.

Discriminant the discriminant of the function \(f(x, y)\) is given by the formula \(D=f_{xx}(x_0,y_0)f_{by}(x_0,y_0)(f_{by}(x_0,y_0))^2\) saddle point given the function \(z=f(x, y), \) the point \((x_0,y_0,f(x_0,y_0))\) is a saddle point if both \(f_x(x_0,y_0)=0\) and \(f_y(x_0,y_0)=0\), but \(f\) does not have a local extreme at \((x_0,y_0)\) Gilbert Strong (MIT) and Edwin “Jed” Herman (Harvey Mud) with many contributing authors.

(Source: www.youtube.com)