Math Problem Statement

Find the absolute maximum and minimum of the function f(x, y) = x^2 + y^2 subject to the constraint x^4 + y^4 = 16.

Solution

To solve the problem, we need to find the absolute maximum and minimum of the function f(x,y)=x2+y2f(x, y) = x^2 + y^2 subject to the constraint x4+y4=16x^4 + y^4 = 16.

Steps to Solve:

  1. Set up the constraint: The constraint is g(x,y)=x4+y416=0g(x, y) = x^4 + y^4 - 16 = 0.

  2. Use the method of Lagrange multipliers: Introduce a multiplier λ\lambda and solve the equations: f=λg\nabla f = \lambda \nabla g where f=(2x,2y)\nabla f = (2x, 2y) and g=(4x3,4y3)\nabla g = (4x^3, 4y^3).

    This gives: 2x=λ4x3,2y=λ4y3.2x = \lambda \cdot 4x^3, \quad 2y = \lambda \cdot 4y^3.

    Simplify: \lambda = \frac{1}{2x^2} \quad \text{(if x \neq 0)}, \quad \lambda = \frac{1}{2y^2} \quad \text{(if y \neq 0)}.

  3. Critical points: From symmetry and the constraint, we test x=0x = 0 or y=0y = 0, and also cases when x=±yx = \pm y. Substitute into the constraint x4+y4=16x^4 + y^4 = 16 to find possible critical points.

  4. Compute f(x,y)f(x, y) at all critical points to find the maximum and minimum values.

Let me calculate these steps and provide the solution.from sympy import symbols, solve, Eq

Define variables

x, y, lambd = symbols('x y lambd')

Function and constraint

f = x2 + y2 # The function g = x4 + y4 - 16 # The constraint

Gradients for Lagrange multipliers

fx = 2 * x fy = 2 * y gx = 4 * x3 gy = 4 * y3

Solve using Lagrange multipliers

eq1 = Eq(fx, lambd * gx) eq2 = Eq(fy, lambd * gy) eq3 = Eq(x4 + y4, 16)

Solve for critical points

critical_points = solve([eq1, eq2, eq3], (x, y, lambd), dict=True)

Evaluate f(x, y) at critical points

results = [] for point in critical_points: x_val = point[x] y_val = point[y] results.append({'x': x_val, 'y': y_val, 'f(x,y)': f.subs({x: x_val, y: y_val})})

Find maximum and minimum values

results_sorted = sorted(results, key=lambda r: r['f(x,y)']) # Sort by f(x, y) results_sorted[{'x': -2**(3/4)I, 'y': -2*(3/4)I, 'f(x,y)': -4sqrt(2)}, {'x': -2**(3/4)I, 'y': 2*(3/4)I, 'f(x,y)': -4sqrt(2)}, {'x': 2**(3/4)I, 'y': -2*(3/4)I, 'f(x,y)': -4sqrt(2)}, {'x': 2**(3/4)I, 'y': 2*(3/4)I, 'f(x,y)': -4sqrt(2)}, {'x': 0, 'y': -2I, 'f(x,y)': -4}, {'x': 0, 'y': 2I, 'f(x,y)': -4}, {'x': -2I, 'y': 0, 'f(x,y)': -4}, {'x': 2I, 'y': 0, 'f(x,y)': -4}, {'x': -2, 'y': 0, 'f(x,y)': 4}, {'x': 0, 'y': -2, 'f(x,y)': 4}, {'x': 0, 'y': 2, 'f(x,y)': 4}, {'x': 2, 'y': 0, 'f(x,y)': 4}, {'x': -2**(3/4), 'y': -2**(3/4), 'f(x,y)': 4sqrt(2)}, {'x': -2**(3/4), 'y': 2**(3/4), 'f(x,y)': 4sqrt(2)}, {'x': 2**(3/4), 'y': -2**(3/4), 'f(x,y)': 4sqrt(2)}, {'x': 2**(3/4), 'y': 2**(3/4), 'f(x,y)': 4sqrt(2)}][{'x': -2**(3/4)I, 'y': -2*(3/4)I, 'f(x,y)': -4sqrt(2)}, {'x': -2**(3/4)I, 'y': 2*(3/4)I, 'f(x,y)': -4sqrt(2)}, {'x': 2**(3/4)I, 'y': -2*(3/4)I, 'f(x,y)': -4sqrt(2)}, {'x': 2**(3/4)I, 'y': 2*(3/4)I, 'f(x,y)': -4sqrt(2)}, {'x': 0, 'y': -2I, 'f(x,y)': -4}, {'x': 0, 'y': 2I, 'f(x,y)': -4}, {'x': -2I, 'y': 0, 'f(x,y)': -4}, {'x': 2I, 'y': 0, 'f(x,y)': -4}, {'x': -2, 'y': 0, 'f(x,y)': 4}, {'x': 0, 'y': -2, 'f(x,y)': 4}, {'x': 0, 'y': 2, 'f(x,y)': 4}, {'x': 2, 'y': 0, 'f(x,y)': 4}, {'x': -2**(3/4), 'y': -2**(3/4), 'f(x,y)': 4sqrt(2)}, {'x': -2**(3/4), 'y': 2**(3/4), 'f(x,y)': 4sqrt(2)}, {'x': 2**(3/4), 'y': -2**(3/4), 'f(x,y)': 4sqrt(2)}, {'x': 2**(3/4), 'y': 2**(3/4), 'f(x,y)': 4sqrt(2)}]success

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Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Lagrange Multipliers
Optimization
Algebra

Formulas

f(x, y) = x^2 + y^2
Constraint: x^4 + y^4 = 16
Lagrange Multipliers: ∇f = λ∇g

Theorems

Lagrange Multipliers
Symmetry in Optimization

Suitable Grade Level

College Level (Calculus III or Advanced Calculus)