Question

Помогите, пожалуйста, срочно!
Исследовать функцию z = z (x, y) на экстремум.

Answer 1

Ответ:

В точке $M_{3}(4;4)$ функция $z$ имеет максимум $z_{max} = 64$

Объяснение:

$z = xy( 12 - x - y)$

$z = xy( 12 - x - y) = 12xy - x^{2}y - y^{2}x$

$z = 12xy - x^{2}y - y^{2}x$

$z'_{x} = (12xy - x^{2}y - y^{2}x)'_{x} = 12y - 2xy - y^{2}$

$z'_{y} = (12xy - x^{2}y - y^{2}x)'_{y} = 12x - 2xy - x^{2}$

$\displaystyle \left \{ {{z_{x}' = 0} \atop {{z_{y}' = 0}} \right.$   $\displaystyle \left \{ {{12y - 2xy - y^{2} =0} \atop {12x - 2xy - x^{2} =0}} \right.$  $\displaystyle \left \{ {{12y - y^{2} =2xy} \atop {12x - x^{2} =2xy}} \right.$  $\displaystyle \left \{ {{ y(12 - y) =2xy} \atop {x(12 - x) =2xy}} \right.$

Пусть x = 0

y(12 - y) =2xy

y(12 - y) =2y * 0

y(12 - y) = 0

$y_{1} = 0$

$12 - y = 0\\y_{2} = 12$

(0;0) , (0;12)

Пусть y = 0

x(12 - x) =2xy

x(12 - x) =2x * 0

x(12 - x) = 0

$x_{1} = 0$

$12 - x = 0\\x_{2} = 12$

(0;0) , (12;0)

$\displaystyle \left \{ {{ y(12 - y) =2xy |:y} \atop {x(12 - x) =2xy|:x}} \right.$ $\displaystyle \left \{ {{12 - y=2x} \atop {12-x=2y}} \right.$  $\displaystyle \left \{ {{y=12 - 2x} \atop {12-x=2y}} \right. \Longrightarrow 12 - x = 2(12 - 2x)$

$12 - x = 2(12 - 2x)$

$12 - x = 24 - 4x$

$3x = 12|:3$

$x = 4$

$y = 12 - 2x = 12 - 2 * 4 = 12 - 8 = 4$

(4;4)

Точки подозрительные на экстремум:

$M_{0} (0;0) , M_{1}(0;12), M_{2}(12;0), M_{3}(4;4)$

$A = z_{xx}''(M)$

$B = z_{xy}''(M)$

$C = z_{yy}''(M)$

$z_{xx}'' = (12y - 2xy - y^{2})_{x}' = -2y$

$z_{xy}'' = (12y - 2xy - y^{2})_{y}' = 12 - 2x -2y$

$z_{yy}'' = (12x - 2xy - x^{2} )_{y}' = - 2x$

$z_{yx}'' = (12x - 2xy - x^{2} )_{x}' = 12 - 2y - 2x$

$M_{1} (0;12) :$

$A = z_{xx}''(M_{1}) = z_{xx}''(0;12) = -2 * 12 = -24$

$B = z_{xy}''(M_{1}) = z_{xy}''(0;12) =$ $12 - 2 * 12 - 2 * 0 = 12 - 24 = -12$

$C = z_{yy}''(M_{1}) = z_{yy}''(0;12) = -2 * 0 = 0$

$\bigtriangleup = AC - B^{2} = -24 * 0 - (-12)^{2} = 0 - 144 = - 144$

Так как $\bigtriangleup < 0$ то в точке $M_{1}(0;12)$ экстремума не существует

$M_{2} (12;0) :$

$A = z_{xx}''(M_{2}) = z_{xx}''(12;0) = -2 * 0 = 0$

$B = z_{xy}''(M_{2}) = z_{xy}''(12;0) =$ $12 - 2 * 12 - 2 * 0 = 12 - 24 = -12$

$C = z_{yy}''(M_{1}) = z_{yy}''(0;12) = -2 * 12 = -24$

$\bigtriangleup = CA - B^{2} = -24 * 0 - (-12)^{2} = 0 - 144 = - 144$

Так как $\bigtriangleup < 0$ то в точке $M_{2}(12;0)$ экстремума не существует

$M_{3} (4;4) :$

$A = z_{xx}''(M_{3}) = z_{xx}''(4;4) = -2 * 4 = -8$

$B = z_{xy}''(M_{3}) = z_{xy}''(4;4) =$ $12 - 2 * 4 - 2 * 4 = 12 - 8 - 8 = 12 - 16 = -4$

$C = z_{yy}''(M_{3}) = z_{yy}''(4;4) = -2 * 4 = -8$

$\bigtriangleup = AC - B^{2} = (-8)*(-8) - (-4)^{2} = 64 - 16 = 48$

Так как $\bigtriangleup > 0$ и $A < 0$ то в точке $M_{3}(4;4)$ максимум функции $z$

$z_{max} = z(M_{3}) = z(4;4) = 4 * 4(12 - 4 - 4) = 16 * (12 -8) = 16* 4 = 64$

$M_{0} (0;0) :$

$A = z_{xx}''(M_{0}) = z_{xx}''(0;0) = -2 * 0 = 0$

$B = z_{xy}''(M_{0}) = z_{xy}''(0;0) =$ $12 - 2 * 0 - 2 * 0 = 12 - 0 = 12$

$C = z_{yy}''(M_{0}) = z_{yy}''(0;0) = -2 * 0 = 0$

$\bigtriangleup = AC - B^{2} = 0 * 0 - (12)^{2} = 0 - 144 = - 144$

Так как $\bigtriangleup < 0$ то в точке $M_{0}(0;0)$ экстремума не существует