Question

Найти экстремум функции z=15xy-x^3-y^3

Answer 1

$z=15xy-x^3-y^3\\ z_x=15y-3x^2\\ z_y=15x-3y^2\\ \left \{ {{15y-3x^2=0} \atop {15x-3y^2}=0} \right. \\ \left \{ {{5y-x^2=0} \atop {5x-y^2}=0} \right. \\ \left \{ {{y= \frac{x^2}{5} } \atop {5x-( \frac{x^2}{5} )^2}=0} \right. \\ \left \{ {{y= \frac{x^2}{5} } \atop {5x-\frac{x^4}{25} }=0} \right. \\ \left \{ {{y= \frac{x^2}{5} } \atop {125x-x^4 }=0} \right. \\ \left \{ {{y= \frac{x^2}{5} } \atop {x(125-x^3) =0} \right. \\ \left \{ {{y= \frac{x^2}{5} } \atop {x=0 \ 125-x^3 =0} \right.$
$\left \{ {{y_1= 0 \ y_2=5 } \atop {x_1=0 \ x_2 =5} \right.$

Стационарные точки
M₁(0,0), M₂(5;5).

$z_{xx}=-6x\\ z_{xy}=15\\ z_{yy}=-6y\\ A_1=-6*0=0\\ B_1=15\\ C_1=-6*0=0\\ \Delta =A_1C_1-B_1^2=0*0-15^2=-225\ \textless \ 0$
В точке M₁(0,0) экстремума нет.

$A_2=-6*5=-30\ \textless \ 0\\ B_2=15\\ C_2=-6*5=-30\ \textless \ 0\\ \Delta =A_2C_2-B_2^2=(-30)*(-30)-15^2=900-225=675\ \textgreater \ 0$
В точке M₂(5,5) максимум.
$z_{max}=15*5*5-5^3-5^3=125$