Respuesta :
Answer:
[tex]Maximum value = \frac{27}{64}[/tex] by using Lagrange multipliers
Step-by-step explanation:
Given the function[tex]f(x, y, z) = x^2y^2z^2[/tex] ……..(1) subject to the condition
Ф(x, y, z) = [tex]x^2+ y^2+ z^2 =4[/tex] ...…..... (2)
Step :1
Form Lagrange's function
F(x ,y, z) = f(x, y, z) + α Ф(x, y, z)
F(x ,y, z) = [tex]x^2y^2z^2 + \alpha (x^2+y^2+z^2)[/tex] ……..(3) here 'α' is the Lagrange's
multipliers
Differentiating partially with respective to 'x' we get and equating zero
[tex]2xy^2z^2 + \alpha (2x) =0[/tex]
on simplification , we get
[tex]\alpha = - y^{2} z^{2}[/tex] .....(a)
Differentiating partially with respective to 'y' we get and equating zero
[tex]x^22yz^2 + \alpha (2y) =0[/tex]
on simplification , we get
[tex]\alpha = - x^{2} z^{2}[/tex] …. (b)
Differentiating partially with respective to 'z' we get and equating zero
[tex]x^2y^22z + \alpha (2z) =0[/tex]
on simplification , we get
[tex]\alpha = - x^{2} y^{2}[/tex] ….. (c)
Step :2
Equating (a) and (b) , we get
[tex]y^{2} z^{2}= x^{2} z^{2}[/tex]
cancellation [tex]z^{2}[/tex] on both sides,we get
[tex]y^{2} = x^{2}[/tex] ….. (d)
Equating (b) and (c) , we get
[tex]x^{2} z^{2}= x^{2} y^{2}[/tex]
cancellation [tex]x^{2}[/tex] on both sides, we get
[tex]z^{2} = y^{2}[/tex] …. (e)
Equating (c) and (a) , we get
[tex]x^{2} y^{2}= y^{2} z^{2}[/tex]
cancellation [tex]y^{2}[/tex] on both sides, we get
[tex]x^{2} = z^{2}[/tex] …. (f)
From (d) , (e), and (f)
[tex]y^{2} = x^{2} = z^{2}[/tex] …. (4)
Using (2) and (4) condition
[tex]x^2+ x^2+ x^2 =4[/tex]
[tex]3x^2 =4[/tex]
[tex]x^2 =\frac{3}{4}[/tex]
and similarly [tex]y^2 =\frac{3}{4}[/tex] and [tex]z^2 =\frac{3}{4}[/tex]
Therefore [tex]x^2 =\frac{3}{4} , y^2 =\frac{3}{4} ,z^2 =\frac{3}{4}[/tex]
The [tex]x = \frac{\sqrt{3} }{2} , y = \frac{\sqrt{3} }{2} ,z = \frac{\sqrt{3} }{2}[/tex]
Step 3:-
Maximum value [tex]f(x, y, z) = x^2y^2z^2[/tex]
[tex]f(x, y, z) = \frac{3X3X3}{4X4X4}[/tex]
[tex]f(x, y, z) = \frac{27}{64}[/tex]
the maximum value = 27 /64
