Math Problem Statement

f(x , y) = e^(x - y) * (x^2 + 2y^2)

Solution

The function f(x,y)=exy(x2+2y2)f(x, y) = e^{x - y} \cdot (x^2 + 2y^2) combines an exponential and a polynomial expression. Let’s analyze and explore it step by step:

Properties and Analysis:

  1. Domain: The function f(x,y)f(x, y) is defined for all real numbers xx and yy because both the exponential exye^{x-y} and the polynomial x2+2y2x^2 + 2y^2 are well-defined for all real x,yx, y.

  2. Partial Derivatives:

    • To compute fx\frac{\partial f}{\partial x}: fx=x(exy(x2+2y2))\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( e^{x-y} \cdot (x^2 + 2y^2) \right) Using the product rule: fx=exyx(x2+2y2)+x(exy)(x2+2y2)\frac{\partial f}{\partial x} = e^{x-y} \cdot \frac{\partial}{\partial x}(x^2 + 2y^2) + \frac{\partial}{\partial x}(e^{x-y}) \cdot (x^2 + 2y^2) Simplifying: fx=exy2x+exy(x2+2y2)\frac{\partial f}{\partial x} = e^{x-y} \cdot 2x + e^{x-y} \cdot (x^2 + 2y^2) fx=exy(2x+x2+2y2)\frac{\partial f}{\partial x} = e^{x-y} \cdot \left(2x + x^2 + 2y^2 \right)

    • To compute fy\frac{\partial f}{\partial y}: fy=y(exy(x2+2y2))\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( e^{x-y} \cdot (x^2 + 2y^2) \right) Using the product rule: fy=exyy(x2+2y2)+y(exy)(x2+2y2)\frac{\partial f}{\partial y} = e^{x-y} \cdot \frac{\partial}{\partial y}(x^2 + 2y^2) + \frac{\partial}{\partial y}(e^{x-y}) \cdot (x^2 + 2y^2) Simplifying: fy=exy4yexy(x2+2y2)\frac{\partial f}{\partial y} = e^{x-y} \cdot 4y - e^{x-y} \cdot (x^2 + 2y^2) fy=exy(4yx22y2)\frac{\partial f}{\partial y} = e^{x-y} \cdot \left(4y - x^2 - 2y^2\right)

  3. Critical Points: To find critical points, solve fx=0\frac{\partial f}{\partial x} = 0 and fy=0\frac{\partial f}{\partial y} = 0 simultaneously.

  4. Growth and Decay:

    • exye^{x-y} grows or decays exponentially depending on the sign of xyx-y.
    • x2+2y2x^2 + 2y^2 grows quadratically, emphasizing the dominance of the yy-axis due to the coefficient 2.
  5. Behavior:

    • f(x,y)f(x, y) will be large in magnitude where xx and yy are large and of similar values (exponential growth from exye^{x-y}).
    • Near the origin (x,y0x, y \to 0), f(0,0)=0f(0, 0) = 0.

Would you like me to analyze further, solve specific problems involving this function, or visualize its graph?


Related Questions:

  1. What is the behavior of f(x,y)f(x, y) as x,yx, y \to \infty?
  2. How can we classify the critical points of f(x,y)f(x, y)?
  3. What is the second partial derivative of f(x,y)f(x, y)?
  4. Can we visualize f(x,y)f(x, y) for specific ranges of xx and yy?
  5. How does f(x,y)f(x, y) behave along the lines x=yx = y or y=0y = 0?

Tip:

When analyzing multivariable functions, always consider symmetry and boundary behavior to simplify the study of critical points and extrema.

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Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Partial Derivatives
Exponential Functions
Polynomial Expressions
Critical Points Analysis

Formulas

f(x, y) = e^(x - y) * (x^2 + 2y^2)
Partial derivatives: ∂f/∂x and ∂f/∂y using the product rule
Critical points: Solve ∂f/∂x = 0 and ∂f/∂y = 0 simultaneously

Theorems

Product Rule for Derivatives
Critical Point Classification

Suitable Grade Level

Undergraduate (Calculus I/II or Multivariable Calculus)