How to Find Relative Extrema Using Single-Variable Substitution Method

Hi there! This is a really clever problem that combines multivariable calculus with a substitution technique. Let's work through it together! 🌟

What We're Solving:

We want to find the relative extrema (local maxima and minima) of the function f(x,y) = [x⁴ + y⁴ - 2(x - y)² - 2] · e^(x⁴ + y⁴ - 2(x - y)² - 2) by using the substitution t = x⁴ + y⁴ - 2(x - y)² - 2 to create a single-variable function g(t) = t · eᵗ.

The Approach:

The brilliant insight here is recognizing that our original function has the form f(x,y) = u(x,y) · e^(u(x,y)) where u(x,y) = x⁴ + y⁴ - 2(x - y)² - 2. This suggests we can study the simpler function g(t) = t · eᵗ first, then understand how extrema of g relate back to extrema of f.

Step-by-Step Solution:

Step 1: Analyze g(t) = t · eᵗ Let's find the critical points of g(t):

• g'(t) = eᵗ + t · eᵗ = eᵗ(1 + t)
• Setting g'(t) = 0: Since eᵗ > 0 always, we need 1 + t = 0, so t = -1

Step 2: Determine the nature of the critical point

• g''(t) = eᵗ(1 + t) + eᵗ = eᵗ(2 + t)
• At t = -1: g''(-1) = e⁻¹(2 - 1) = e⁻¹ > 0
• Since g''(-1) > 0, t = -1 gives a local minimum of g(t)
• The minimum value is g(-1) = (-1) · e⁻¹ = -1/e

Step 3: Connect back to f(x,y) Now we need points (x,y) where: x⁴ + y⁴ - 2(x - y)² - 2 = -1

Simplifying: x⁴ + y⁴ - 2(x - y)² = 1

Step 4: Expand and solve x⁴ + y⁴ - 2(x² - 2xy + y²) = 1 x⁴ + y⁴ - 2x² + 4xy - 2y² = 1

This is challenging to solve directly, but we can look for symmetric solutions. Try x = y: x⁴ + x⁴ - 2(0)² = 1 2x⁴ = 1 x⁴ = 1/2 x = ±(1/2)^(1/4)

So we get critical points at (±(1/2)^(1/4), ±(1/2)^(1/4)).

Step 5: Verify these are extrema of f(x,y) Since g(t) has its unique minimum at t = -1, and our function f(x,y) = g(u(x,y)) where u(x,y) = t, the extrema of f occur where:

• 1. u(x,y) = -1 (giving the value f = -1/e), OR
• 2. At critical points of u(x,y) where ∇u = 0

The Answer:

The function f(x,y) has relative minima at the points where x⁴ + y⁴ - 2(x - y)² - 2 = -1, including the symmetric points (±(1/2)^(1/4), ±(1/2)^(1/4)). At these points, f(x,y) = -1/e ≈ -0.368.

Memory Tip:

When you see a function of the form h(u(x,y)) where h is simpler than the original multivariable function, try analyzing h first! The extrema often correspond beautifully between the single-variable and multivariable cases. It's like finding the "skeleton" of the problem! 🦴

Great job tackling this sophisticated problem - the substitution technique is a powerful tool in your calculus toolkit!