Answer:
Step-by-step explanation:
Given that x+y+z = 48
Sum of squares [tex]x^2+y^2+z^2[/tex]
is minimum
Let us eliminate one variable say z = 48-x-y
f(x,y) = [tex]x^2+y^2+z^2\\=x^2+y^2+(48-x-y)^2\\= 2x^2+2y^2 +48^2-96x-96y-2xy[/tex]
Use partial derivatives to find min of f
[tex]f_x = 4x-2(48-x-y)\\f_y = 4y-2(48-x-y)\\f_xx = 6: f_yy = 6y\\f_{xy} =2=f_yx[/tex]
Equate I derivative to 0
2x+6y =96 and 6x+2y = 96
Solving we get
x=y =12
z=48-24 = 24
[tex]D^2 = 48*48-16>0\\f_{xx} (12,12) >0[/tex]
so minimum when x=y =12 and z = 24
