f ( x , y ) = x 3 y 2 ( 24 − x − y ) = 24 x 3 y 2 − x 4 y 2 − x 3 y 3 f(x,y) = x^3y^2(24-x-y) = 24x^3y^2-x^4y^2-x^3y^3 f ( x , y ) = x 3 y 2 ( 24 − x − y ) = 24 x 3 y 2 − x 4 y 2 − x 3 y 3
necessary extremum condition:
Let M 0 ( x 0 , y 0 ) M_0(x_0, y_0) M 0 ( x 0 , y 0 )
{ ∂ f ∂ x ( x 0 , y 0 ) = 0 ∂ f ∂ y ( x 0 , y 0 ) = 0 \begin{cases}
\cfrac{\partial f}{\partial x}(x_0,y_0) = 0\\\cfrac{\partial f}{\partial y}(x_0,y_0) = 0
\end{cases} ⎩ ⎨ ⎧ ∂ x ∂ f ( x 0 , y 0 ) = 0 ∂ y ∂ f ( x 0 , y 0 ) = 0
Find first partial derivatives:
∂ f ∂ x = 72 x 2 y 2 − 4 x 3 y 2 − 3 x 2 y 3 ∂ f ∂ y = { ∂ f ∂ x = 72 x 2 y 2 − 4 x 3 y 2 − 3 x 2 y 3 = 0 ∂ f ∂ y = 48 x 3 y − 2 x 4 y − 3 x 3 y 2 = 0 \cfrac{\partial f}{\partial x} = 72x^2y^2 - 4x^3y^2-3x^2y^3 \\
\cfrac{\partial f}{\partial y} = \\
\begin{cases}
\cfrac{\partial f}{\partial x} = 72x^2y^2 - 4x^3y^2-3x^2y^3 = 0 \\
\cfrac{\partial f}{\partial y} = 48x^3y-2x^4y-3x^3y^2=0 \\
\end{cases} \\ ∂ x ∂ f = 72 x 2 y 2 − 4 x 3 y 2 − 3 x 2 y 3 ∂ y ∂ f = ⎩ ⎨ ⎧ ∂ x ∂ f = 72 x 2 y 2 − 4 x 3 y 2 − 3 x 2 y 3 = 0 ∂ y ∂ f = 48 x 3 y − 2 x 4 y − 3 x 3 y 2 = 0
Solve this system for x , y x,y x , y
{ x 2 y 2 ( 72 − 4 x − 3 y ) = 0 x 2 y ( 48 x − 2 x 2 − 3 y ) = 0 { x 2 y 2 = 0 o r − 4 x − 3 y + 72 = 0 x 2 y = 0 o r − 2 x 2 + 48 x − 3 y = 0 \begin{cases}
x^2y^2(72-4x-3y) = 0\\
x^2y(48x-2x^2-3y) = 0
\end{cases}\\
\begin{cases}
x^2y^2 = 0 \,or\,-4x-3y+72 = 0\\
x^2y =0\, or\,-2x^2+48x-3y=0
\end{cases} { x 2 y 2 ( 72 − 4 x − 3 y ) = 0 x 2 y ( 48 x − 2 x 2 − 3 y ) = 0 { x 2 y 2 = 0 or − 4 x − 3 y + 72 = 0 x 2 y = 0 or − 2 x 2 + 48 x − 3 y = 0
Find the roots of first line:
x 2 y 2 = 0 o r − 4 x − 3 y + 72 = 0 x^2y^2 = 0 \,or\,-4x-3y+72 = 0 x 2 y 2 = 0 or − 4 x − 3 y + 72 = 0
From second equation:
y = 24 − 4 3 x y = 24 - \cfrac{4}{3}x y = 24 − 3 4 x
And put it in first equation:
24 x − 4 3 x 2 = 0 − 4 x 2 + 72 x = 0 x ( − 4 x + 72 ) = 0 x = 0 o r x = 18 y = 24 o r y = 0 24x-\cfrac{4}{3}x^2 = 0\\
-4x^2+72x=0\\
x(-4x+72) =0\\
x=0\;or \; x =18\\
y=24\; or\;y=0 24 x − 3 4 x 2 = 0 − 4 x 2 + 72 x = 0 x ( − 4 x + 72 ) = 0 x = 0 or x = 18 y = 24 or y = 0
M 1 ( 0 ; 24 ) , M 2 ( 18 ; 0 ) M_1(0;24),M_2(18;0) M 1 ( 0 ; 24 ) , M 2 ( 18 ; 0 )
Find the roots of second line:
x 2 y = 0 o r ( 48 x − 2 x 2 − 3 y ) = 0 x^2y=0\, or\,(48x-2x^2-3y) = 0 x 2 y = 0 or ( 48 x − 2 x 2 − 3 y ) = 0
From second equation:
y = − 2 3 x 2 + 16 x y = -\cfrac{2}{3}x^2+16x y = − 3 2 x 2 + 16 x
And put it in first equation:
x 2 ( − 2 3 x 2 + 16 x ) = 0 x 3 ( − 2 3 x + 16 ) = 0 x = 0 o r x = 24 y = 0 o r y = 0 M 3 ( 0 ; 0 ) , M 4 ( 24 ; 0 ) x^2(-\cfrac{2}{3}x^2+16x)=0\\
x^3(-\cfrac{2}{3}x+16)= 0\\
x = 0\;or\; x =24\\
y=0\; or \;y=0\\
M_3(0;0), M_4(24;0) x 2 ( − 3 2 x 2 + 16 x ) = 0 x 3 ( − 3 2 x + 16 ) = 0 x = 0 or x = 24 y = 0 or y = 0 M 3 ( 0 ; 0 ) , M 4 ( 24 ; 0 )
{ M 1 ( 0 ; 24 ) , M 2 ( 18 ; 0 ) M 3 ( 0 ; 0 ) , M 4 ( 24 ; 0 ) \begin{cases}
M_1(0;24),M_2(18;0)\\
M_3(0;0), M_4(24;0)
\end{cases} { M 1 ( 0 ; 24 ) , M 2 ( 18 ; 0 ) M 3 ( 0 ; 0 ) , M 4 ( 24 ; 0 )
The obtained points are the solution of two aggregates of equations, but the necessary extremum condition says that the points must be a solution to the system. Therefore, the system has one solution - M 3 ( 0 ; 0 ) M_3(0;0) M 3 ( 0 ; 0 )
If ∂ 2 f ∂ x 2 ( x 0 ; y 0 ) = A , ∂ 2 f ∂ y 2 ( x 0 ; y 0 ) = B , ∂ 2 f ∂ x ∂ y ( x 0 ; y 0 ) = C \cfrac{\partial^2 f}{\partial x^2}(x_0;y_0) = A, \cfrac{\partial^2 f}{\partial y^2}(x_0;y_0) = B, \cfrac{\partial^2 f}{\partial x \partial y}(x_0;y_0) = C ∂ x 2 ∂ 2 f ( x 0 ; y 0 ) = A , ∂ y 2 ∂ 2 f ( x 0 ; y 0 ) = B , ∂ x ∂ y ∂ 2 f ( x 0 ; y 0 ) = C
then :
if A C − B 2 > 0 a n d A > 0 ( A < 0 ) , t h e n M ( x 0 ; y 0 ) − m i n i m u m ( m a x i m u m ) AC - B^2 > 0\, and\, A>0 \,(A<0), then\, M(x_0;y_0) - minimum (maximum) A C − B 2 > 0 an d A > 0 ( A < 0 ) , t h e n M ( x 0 ; y 0 ) − minim u m ( ma x im u m )
If A C − B 2 < 0 , t h e n M ( x 0 ; y 0 ) − i s n ′ t e x t r e m u m AC-B^2 < 0, then\, M(x_0;y_0) -isn't\; extremum A C − B 2 < 0 , t h e n M ( x 0 ; y 0 ) − i s n ′ t e x t re m u m
if A C − B 2 = 0 , t h e n w e c a n ′ t s a y a n y t h i n g . AC-B^2 = 0, \; then \, we \,can't\, say \,anything. A C − B 2 = 0 , t h e n w e c a n ′ t s a y an y t hin g .
∂ 2 f ∂ x 2 = 144 x y 2 − 12 x 2 y 2 − 6 x y 3 ∂ 2 f ∂ y 2 = 48 x 3 − 2 x 4 − 6 x 3 y ∂ 2 f ∂ x ∂ y = 144 x 2 y − 8 x 3 y − 49 x 2 y 2 \cfrac{\partial^2 f}{\partial x^2} = 144xy^2 -12x^2y^2-6xy^3\\
\cfrac{\partial^2 f}{\partial y^2} = 48x^3-2x^4-6x^3y\\
\cfrac{\partial^2 f}{\partial x \partial y} = 144x^2y - 8x^3y-49x^2y^2 ∂ x 2 ∂ 2 f = 144 x y 2 − 12 x 2 y 2 − 6 x y 3 ∂ y 2 ∂ 2 f = 48 x 3 − 2 x 4 − 6 x 3 y ∂ x ∂ y ∂ 2 f = 144 x 2 y − 8 x 3 y − 49 x 2 y 2
A ( 0 ; 0 ) = 0 B ( 0 ; 0 ) = 0 , C ( 0 ; 0 ) = 0 A(0;0) = 0\;B(0;0) = 0, \;C(0;0) = 0 A ( 0 ; 0 ) = 0 B ( 0 ; 0 ) = 0 , C ( 0 ; 0 ) = 0
A C − B 2 = 0 AC-B^2 = 0 A C − B 2 = 0 M 3 ( 0 ; 0 ) M_3(0;0) M 3 ( 0 ; 0 )
#339914 on Dec 2023
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