Находим частные производные:
[m]\frac{ ∂ Φ }{ ∂x }=(2x^2+12xy+y^2+ λ (x^2+4y^2-25))`_{x}=4x+12y+2x λ [/m]
[m]\frac{ ∂ Φ }{ ∂y }=(2x^2+12xy+y^2)`_{y}=12x+2y+9y λ [/m]
[m]\frac{ ∂ Φ }{ ∂ λ }=(2x^2+12xy+y^2)`_{y}=x^2+4y^2-25[/m]
Находим стационарные точки:
[m]\left\{\begin {matrix}\frac{ ∂z }{ ∂x }=0\\\frac{ ∂z }{ ∂y }=0\\\frac{ ∂ Φ }{ ∂ λ }=0\end {matrix}\right.[/m]
⇒ [m]\left\{\begin {matrix}4x+12y+2x λ=0\\12x+2y+8y λ=0\\x^2+4y^2-25=0\end {matrix}\right.[/m] ⇒ [m]\left\{\begin {matrix} λ=\frac{-4x-12y}{2x}\\λ=\frac{-12x-2y}{8y}\\x^2+4y^2-25=0\end {matrix}\right.[/m]⇒ [m]\left\{\begin {matrix} λ=\frac{-4x-12y}{2x}\\\frac{-4x-12y}{2x}=\frac{-12x-2y}{8y}\\x^2+4y^2-25=0\end {matrix}\right.[/m] ⇒ [m]\left\{\begin {matrix} λ=\frac{-4x-12y}{2x}\\-2-6\frac{y}{x}=-\frac{6}{4}\frac{x}{y-\frac{1}{4}}\\x^2+4y^2-25=0\end {matrix}\right.[/m] ⇒
[m]\left\{\begin {matrix} λ=\frac{-4x-12y}{2x}\\\frac{y}{x}=-16\\x^2+4(-16x)^2-25=0\end {matrix}\right.[/m] или [m]\left\{\begin {matrix} λ=\frac{-4x-12y}{2x}\\\frac{y}{x}=9\\x^2+4(9x)^2-25=0\end {matrix}\right.[/m]
[m]\left\{\begin {matrix} λ=\frac{-4x-12y}{2x}\\\frac{y}{x}=-16\\1025x^2=25\end {matrix}\right.[/m] или [m]\left\{\begin {matrix} λ=\frac{-4x-12y}{2x}\\\frac{y}{x}=9\\325x^2=25\end {matrix}\right.[/m]
[m]\left\{\begin {matrix} λ=\frac{-4x-12y}{2x}\\\frac{y}{x}=-16\\401x^2=1\end {matrix}\right.[/m] или [m]\left\{\begin {matrix} λ=\frac{-4x-12y}{2x}\\\frac{y}{x}=9\\13x^2=1\end {matrix}\right.[/m]
Получим 4 точки.
Исследовать их на максимум, минимум.