Question

Найти экстремум функции:
z=2x^3+5y^2-5xy+42


Срочно сыну! =)

Answer 1

Ответ:

Функция имеет локальный минимум в точке $\bigg(\dfrac{5}{12} ; \dfrac{5}{24} \bigg)$ равный

$z_{min} = \dfrac{72451}{1728}$

Пошаговое объяснение:

$z = 2x^{3} + 5y^{2} - 5xy + 42$

$\displaystyle \frac{\partial z}{\partial x} = \frac{\partial}{\partial x} \bigg( 2x^{3} + 5y^{2} - 5xy + 42 \bigg) = 6x^{2} - 5y$

$\displaystyle \frac{\partial^{2} z}{\partial x^{2}} = \frac{\partial}{\partial x} \bigg( \frac{\partial z}{\partial x} \bigg) = \frac{\partial}{\partial x} \bigg( 6x^{2} - 5y \bigg) = 12x$

$\displaystyle \frac{\partial z}{\partial y} = \frac{\partial}{\partial y} \bigg( 2x^{3} + 5y^{2} - 5xy + 42 \bigg) = 10y - 5x$

$\displaystyle \frac{\partial^{2} z}{\partial y^{2}} = \frac{\partial}{\partial y} \bigg( \frac{\partial z}{\partial y} \bigg) = \frac{\partial}{\partial y} \bigg( 10y - 5x \bigg) = 10$

$\displaystyle \frac{\partial^{2} z}{\partial x \partial y} = \frac{\partial}{\partial x} \bigg( \frac{\partial z}{\partial y} \bigg) = \frac{\partial}{\partial x} \bigg( 10y - 5x \bigg) = -5$

$\displaystyle \left \{ {{\dfrac{\partial z}{\partial x}=0} \atop {\dfrac{\partial z}{\partial y}=0}} \right \ \ \left \{ {{6x^{2} - 5y = 0} \atop {10y - 5x = 0}} \right \ \left \{ {{6x^{2} = 5y} \atop {10y = 5x|:5}} \right$

$\displaystyle \left \{ {{x = 2y} \atop {6x^{2} = 5y}} \right \Longrightarrow 6 \cdot (2y)^{2} = 5y \Longleftrightarrow 24y^{2} = 5y$

$24y^{2} = 5y$

$24y^{2} - 5y =0$

$y(24y - 5) = 0$

$y = 0; \ \ \ 24y - 5 = 0$

$y = 0; \ \ \ 24y = 5 |:24$

$y_{1} = 0; \ \ \ y_{2} = \dfrac{5}{24}$

$x_{1} = 2y_{1} = 2 \cdot 0 = 0$

$x_{2} = 2y_{2} = 2 \cdot \dfrac{5}{24} = \dfrac{5}{12}$

$\bigg( 0; 0 \bigg), \bigg(\dfrac{5}{12} ; \dfrac{5}{24} \bigg)$

1) $(0;0)$

$A = \displaystyle \frac{\partial^{2} z}{\partial x^{2}} \bigg |_{(0;0)} = 12 \cdot 0 = 0$

$B = \displaystyle \frac{\partial^{2} z}{\partial x \partial y} \bigg |_{(0;0)} = -5$

$C = \displaystyle \frac{\partial^{2} z}{\partial y^{2}} \bigg |_{(0;0)} =10$

$\Delta = AC - B^{2} = 10 \cdot 0 - (-5)^{2} = -25 < 0$

Так как $\Delta < 0$, то в данной точке функции экстремума не имеет.

2) $\bigg(\dfrac{5}{12} ; \dfrac{5}{24} \bigg)$

$A = \displaystyle \frac{\partial^{2} z}{\partial x^{2}} \Bigg |_{\bigg(\dfrac{5}{12} ; \dfrac{5}{24} \bigg)} = 12 \cdot \dfrac{5}{12} = 5$

$B = \displaystyle \frac{\partial^{2} z}{\partial x \partial y} \Bigg |_{\bigg(\dfrac{5}{12} ; \dfrac{5}{24} \bigg)} = -5$

$C = \displaystyle \frac{\partial^{2} z}{\partial y^{2}} \Bigg |_{\bigg(\dfrac{5}{12} ; \dfrac{5}{24} \bigg)} =10$

$\Delta = AC - B^{2} = 10 \cdot 5 - (-5)^{2} = 50 -25 = 25 > 0$

Так в данной точке $\Delta > 0$ и $\displaystyle \frac{\partial^{2} z}{\partial x^{2}} \Bigg |_{\bigg(\dfrac{5}{12} ; \dfrac{5}{24} \bigg)} > 0$, то точка  $\bigg(\dfrac{5}{12} ; \dfrac{5}{24} \bigg)$ является точкой локального минимума.

Минимум функции:

$z_{min} = z\bigg(\dfrac{5}{12} ; \dfrac{5}{24} \bigg) = \dfrac{72451}{1728}$