Answer:
Step-by-step explanation:
[tex]x + 2y + 3z = 12[/tex]
[tex]z=\frac{12-x-2y}{3}[/tex]
Volume = [tex]V=f(x,y) = xy(\frac{12-x-2y}{3} )[/tex]
find partial derivatives using product rule
[tex]f_x =\frac{y}{3} (12-2x-2y)\\f_y = \frac{x}{3} (12-x-4y)[/tex]
i.e.
Using maximum for partial derivatives, we equate first partial derivative to 0.
y=0 or x+y =6
x=0 or x+4y =12
Simplify to get y =2, x = 4
thus critical points are (4,2) (6,0) (0,3)
Of these D the II derivative test gives
D<0 only for (4,2)
Hence maximum volume is when x=4, y=2, z= 4/3
Max volume is = 4(2)(4/3) = 32/3
