Math Problem Statement

Per trovare i punti critici, poniamo entrambe le derivate uguali a zero:

2x1+x2+y2=0⇒x=01+x2+y22x​=0⇒x=02y1+x2+y2=0⇒y=01+x2+y22y​=0⇒y=0

Quindi, l'unico punto critico è (x,y)=(0,0)(x,y)=(0,0). spiega meglio sto passsaggio

Solution

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Math Problem Analysis

Mathematical Concepts

Calculus
Partial Derivatives
Critical Points

Formulas

Partial derivative of f(x, y) with respect to x: ∂f/∂x = 2x + x_2 + y^2
Partial derivative of f(x, y) with respect to y: ∂f/∂y = 2y + x_2 + y^2

Theorems

Conditions for critical points (partial derivatives equal to zero)

Suitable Grade Level

College-level Calculus