The profit in dollars in producing x- items of some commodity is given by the equation P = - 40 x^2 + 1000 x - 5250 .

Question

How many items should be produced to maximize the profit?
What is the maximum profit?

Answer 1

do you know Calculus?
If so, then dP/dx = -80x + 1000
= 0 for a max of P
80x = 1000
x = 1000/80 = 12.5

I assume you can't make a partial product , so either make 12 or 13 to get the same maximum profit

If you don't know calculus ....
the x value of the vertex is -b/(2a)
= -1000/(2(-40)) = 12.5
..... same as above

or

complete the square:
P = -40(x^2 - 25x + ....) - 5250
= -40(x^2 - 25x + 156.25 - 156.25) - 5250
= -40( (x-12.5)^2 - 156.25) - 5250
= -40(x-12.5)^2 + 6250 - 5250
= -40(x-12.5)^2 + 1000
so for a max of P, x = 12.5, but as I said above, we can't make partial items
so either x = 12 or x = 13

( I would make 12. Why make 13 when 12 gives us the same profit ? )