Calculus
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Boundless Calculus
Advanced Topics in Single-Variable Calculus and an Introduction to Multivariable Calculus
Partial Derivatives
Calculus Textbooks Boundless Calculus Advanced Topics in Single-Variable Calculus and an Introduction to Multivariable Calculus Partial Derivatives
Calculus Textbooks Boundless Calculus Advanced Topics in Single-Variable Calculus and an Introduction to Multivariable Calculus
Calculus Textbooks Boundless Calculus
Calculus Textbooks
Calculus
Concept Version 6
Created by Boundless

Applications of Minima and Maxima in Functions of Two Variables

Finding extrema can be a challenge with regard to multivariable functions, requiring careful calculation.

Learning Objective

  • Identify steps necessary to find the minimum and maximum in multivariable functions


Key Points

    • The second derivative test is a criterion for determining whether a given critical point of a real function of one variable is a local maximum or a local minimum using the value of the second derivative at the point.
    • To find minima/maxima for functions with two variables, we must first find the first partial derivatives with respect to $x$ and $y$ of the function.
    • The function $z = f(x, y) = (x+y)(xy + xy^2)$ has saddle points at $(0,-1)$ and $(1,-1)$ and a local maximum at $\left(\frac{3}{8}, -3.4\right)$.

Terms

  • critical point

    a maximum, minimum, or point of inflection on a curve; a point at which the derivative of a function is zero or undefined

  • multivariable

    concerning more than one variable


Full Text

We have learned how to find the minimum and maximum in multivariable functions. As previously mentioned, finding extrema can be a challenge with regard to multivariable functions. In particular, we learned about the second derivative test, which is a criterion for determining whether a given critical point of a real function of one variable is a local maximum or a local minimum, using the value of the second derivative at the point. In this atom, we will find extrema for a function with two variables.

Example

Find and label the critical points of the following function: 

$z = f(x, y) = (x+y)(xy + xy^2)$

Plot of $z = (x+y)(xy+xy^2)$

The maxima and minima of this plot cannot be found without extensive calculation.

To solve this problem we must first find the first partial derivatives of the function with respect to $x$ and $y$:

$\displaystyle{\frac{\partial z}{\partial x} = y(2x +y)(y+1)}$

$\displaystyle{\frac{\partial z}{\partial y} = x \left( 3y^2 +2y(x+1) + x \right)}$

Looking at $\frac{\partial z}{\partial x} = 0$, we see that $y$ must equal $0$, $-1$, or $-2x$.

We plug the first solution, $y=0$, into the next equation, and get:

$\displaystyle{\frac{\partial z}{\partial y} = x \left( 3y^2 +2y(x+1) + x \right)\\ \,\quad = x^2}$

There were other possibilities for $y$, so for $y=-1$ we have:

$\displaystyle{\frac{\partial z}{\partial y} = x \left( 3 -2(x+1) + x \right) \\ \,\quad= x(1-x)\\ \,\quad= 0}$

So $x$ must be equal to $1$ or $0$. Finally, for $y=-2x$:

$\displaystyle{\frac{\partial z}{\partial y} = x \left( 3(-2x)^2 +2(-2x)(x+1) + x \right) \\ \,\quad= x^2(8x-3) \\ \,\quad= 0}$

So $x$ must equal $0$ or  for $y = 0$ and $y = -\frac{3}{4}$, respectively.

Let's list all the critical values now:

$\displaystyle{(x,y) \in {(0,0), (0, -1), (1,-1), \left(\frac{3}{8}, -\frac{3}{4}\right)}}$

Now we have to label the critical values using the second derivative test. Plugging in all the different critical values we found to label them, we have:

  • $D(0, 0) = 0$
  • $D(0, -1) = -1$
  • $D(1, -1) = -1$
  • $D\left(\frac{3}{8}, -\frac{3}{4}\right) = 0.210938$

We can now label some of the points:

  • at (0, −1), $f(x, y)$ has a saddle point
  • at (1, −1), $f(x, y)$ has a saddle point
  • at $\left(\frac{3}{8}, -\frac{3}{4}\right)$ $f(x, y)$ has a local maximum, since $f_{xx} = -\frac{3}{8} < 0$

At the remaining point we need higher-order tests to find out what exactly the function is doing.

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