Section 12.8 - Maximum/Minimum Problems **Basic Skills**9–18. Critical points: *Find all critical points of the following functions.* ### Problem 18: Multivariable FunctionConsider the function $ f(x, y) = e^{x^2 y^2 - 2xy^2 + y^2} $.#### Explanation:This function is a multivariable exponential function where the exponent is a quadratic expression in terms of $x$ and $y$. The expression in the exponent, $x^2y^2 - 2xy^2 + y^2$, suggests interactions between the variables through their products and powers, indicating potential surface interactions or geometric shapes when plotted in three dimensions.- **Quadratic Terms:** - $x^2y^2$: A term that implies interaction between $x$ and $y$ due to the multiplication of their squares. - $-2xy^2$: A mixed term that affects the shape, shifting it due to the negative factor. - $y^2$: Adds a vertical quadratic effect, influencing the $y$-axis symmetry.This function might be used in applications involving bivariate data behavior, such as modeling surfaces or potential fields.

Answered: 18. f (z,y) = e²³y²–2zy²+y?

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Section 12.8 - Maximum/Minimum Problems

**Basic Skills**

9–18. Critical points: *Find all critical points of the following functions.*

### Problem 18: Multivariable Function

Consider the function $ f(x, y) = e^{x^2 y^2 - 2xy^2 + y^2} $.

#### Explanation:
This function is a multivariable exponential function where the exponent is a quadratic expression in terms of $x$ and $y$. The expression in the exponent, $x^2y^2 - 2xy^2 + y^2$, suggests interactions between the variables through their products and powers, indicating potential surface interactions or geometric shapes when plotted in three dimensions.

- **Quadratic Terms:**
- $x^2y^2$: A term that implies interaction between $x$ and $y$ due to the multiplication of their squares.
- $-2xy^2$: A mixed term that affects the shape, shifting it due to the negative factor.
- $y^2$: Adds a vertical quadratic effect, influencing the $y$-axis symmetry.

This function might be used in applications involving bivariate data behavior, such as modeling surfaces or potential fields.

Expert Solution

Step 1

The function is given as,

$f (x, y) = e^{x^{2} y^{2} - 2 x y^{2} + y^{2}}$

Differentiate w.r.t. x,

$\begin{array}{rcl} \frac{\partial f}{\partial x} & = & \frac{\partial}{\partial x} (e^{x^{2} y^{2} - 2 x y^{2} + y^{2}}) \\ = & \frac{1}{e^{x^{2} y^{2} - 2 x y^{2} + y^{2}}} \cdot (2 x y^{2} - 2 y^{2}) \\ = & \frac{2 x y^{2} - 2 y^{2}}{e^{x^{2} y^{2} - 2 x y^{2} + y^{2}}} \end{array}$

Differentiate w.r.t. y,

$\begin{array}{rcl} \frac{\partial f}{\partial y} & = & \frac{\partial}{\partial y} (e^{x^{2} y^{2} - 2 x y^{2} + y^{2}}) \\ = & \frac{1}{e^{x^{2} y^{2} - 2 x y^{2} + y^{2}}} \cdot (2 x^{2} y - 4 x y + 2 y) \\ = & \frac{2 x^{2} y - 4 x y + 2 y}{e^{x^{2} y^{2} - 2 x y^{2} + y^{2}}} \end{array}$

Step 2

The critical points are the points obtained by solving the following equations,

$\frac{\partial f}{\partial x} = 0$
$\frac{\partial f}{\partial y} = 0$

Solve first equation,

$\begin{array}{rcl} \frac{2 x y^{2} - 2 y^{2}}{e^{x^{2} y^{2} - 2 x y^{2} + y^{2}}} \end{array} = 0 \Rightarrow \begin{array}{rcl} 2 \end{array} \begin{array}{rcl} x \end{array} \begin{array}{rcl} y \end{array} \begin{array}{rcl} ^{2} \end{array} \begin{array}{rcl} - \end{array} \begin{array}{rcl} 2 \end{array} \begin{array}{rcl} y \end{array} \begin{array}{rcl} ^{2} \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} 0 \end{array} \begin{array}{rcl} \Rightarrow \end{array} \begin{array}{rcl} 2 \end{array} \begin{array}{rcl} y \end{array} \begin{array}{rcl} ^{2} \end{array} \begin{array}{rcl} (x - 1) \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} 0 \end{array} \begin{array}{rcl} \Rightarrow \end{array} \begin{array}{rcl} x \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} 1 \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} o \end{array} \begin{array}{rcl} r \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} y \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} 0 \end{array}$

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