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4.7 Maxima/minima problems  (Page 11/10)

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  • Use partial derivatives to locate critical points for a function of two variables.
  • Apply a second derivative test to identify a critical point as a local maximum, local minimum, or saddle point for a function of two variables.
  • 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.

Critical points

For functions of a single variable, we defined critical points as the values of the function when the 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.

Definition

Let z = f ( x , y ) be a function of two variables that is differentiable on an open set containing the point ( x 0 , y 0 ) . The point ( x 0 , y 0 ) is called a critical point of a function of two variables     f if one of the two following conditions holds:

  1. f x ( x 0 , y 0 ) = f y ( x 0 , y 0 ) = 0
  2. Either f x ( x 0 , y 0 ) or f y ( x 0 , y 0 ) does not exist.

Finding critical points

Find the critical points of each of the following functions:

  1. f ( x , y ) = 4 y 2 − 9 x 2 + 24 y + 36 x + 36
  2. g ( x , y ) = x 2 + 2 x y − 4 y 2 + 4 x − 6 y + 4
  1. First, we calculate f x ( x , y ) and f y ( x , y ) :
    f x ( x , y ) = 1 2 ( −18 x + 36 ) ( 4 y 2 − 9 x 2 + 24 y + 36 x + 36 ) −1 / 2 = −9 x + 18 4 y 2 − 9 x 2 + 24 y + 36 x + 36 f y ( x , y ) = 1 2 ( 8 y + 24 ) ( 4 y 2 − 9 x 2 + 24 y + 36 x + 36 ) −1 / 2 = 4 y + 12 4 y 2 − 9 x 2 + 24 y + 36 x + 36 .

    Next, we set each of these expressions equal to zero:
    −9 x + 18 4 y 2 − 9 x 2 + 24 y + 36 x + 36 = 0 4 y + 12 4 y 2 − 9 x 2 + 24 y + 36 x + 36 = 0 .

    Then, multiply each equation by its common denominator:
    − 9 x + 18 = 0 4 y + 12 = 0 .

    Therefore, x = 2 and y = −3 , so ( 2 , −3 ) is a critical point of f .
    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. Since the denominator is the same in each partial derivative, we need only do this once:
    4 y 2 − 9 x 2 + 24 y + 36 x + 36 = 0 .

    This equation represents a hyperbola. We should also note that the domain of f consists of points satisfying the inequality
    4 y 2 − 9 x 2 + 24 y + 36 x + 36 ≥ 0 .

    Therefore, any points on the hyperbola are not only critical points, they are also on the boundary of the domain. To put the hyperbola in standard form, we use the method of completing the square:
    4 y 2 − 9 x 2 + 24 y + 36 x + 36 = 0 4 y 2 − 9 x 2 + 24 y + 36 x = −36 4 y 2 + 24 y − 9 x 2 + 36 x = −36 4 ( y 2 + 6 y ) − 9 ( x 2 − 4 x ) = −36 4 ( y 2 + 6 y + 9 ) − 9 ( x 2 − 4 x + 4 ) = −36 + 36 − 36 4 ( y + 3 ) 2 − 9 ( x − 2 ) 2 = −36 .

    Dividing both sides by −36 puts the equation in standard form:
    4 ( y + 3 ) 2 −36 − 9 ( x − 2 ) 2 −36 = 1 ( x − 2 ) 2 4 − ( y + 3 ) 2 9 = 1 .

    Notice that point ( 2 , −3 ) is the center of the hyperbola.
  2. First, we calculate g x ( x , y ) and g y ( x , y ) :
    g x ( x , y ) = 2 x + 2 y + 4 g y ( x , y ) = 2 x − 8 y − 6 .

    Next, we set each of these expressions equal to zero, which gives a system of equations in x and y :
    2 x + 2 y + 4 = 0 2 x − 8 y − 6 = 0 .

    Subtracting the second equation from the first gives 10 y + 10 = 0 , so y = −1 . Substituting this into the first equation gives 2 x + 2 ( −1 ) + 4 = 0 , so x = −1 . Therefore ( −1 , −1 ) is a critical point of g ( [link] ). There are no points in ℝ 2 that make either partial derivative not exist.
    The function g(x, y) = x2 + 2xy – 4y2 + 4x – 7y + 4 is shown with critical point (–1, –1, 6). The critical point is located where the derivative in the x and y directions are both zero.
    The function g ( x , y ) has a critical point at ( −1 , −1 , 6 ) .
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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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