for max Q, we need
∂Q/∂x=0 and ∂Q∂y=0
∂Q/∂x = 4x
∂Q/∂y = 2y+6
so, there's a max/min at (0,-3).
Since the 2nd partials are both positive, it's a minimum.
So, we need to maximize
2x^2+y^2+6y subject to
x+y = 120
Q(0,120) = 15120
Q(120,0) = 28800
Q(119,1)
Clearly, since 2x^2 decreases faster than y^2+6y increases for each dollar moved from x to y, the max is Q(120,0).
Doesn't seem right, since some equipment must be used.
If x thousand dollars is spent on labor and y thousand dollars is spent on euqpment,The output of a certain factory will be Q(x,y)=2x^2+y^2+6y units.If 120,000 dollars is available,how should this be allocated between labor and equipment
to generate the largest possible output
1 answer