Asked by Lucy
The profit in dollars in producing x- items of some commodity is given by the equation P = - 40 x^2 + 1000 x - 5250 .
How many items should be produced to maximize the profit?
What is the maximum profit?
How many items should be produced to maximize the profit?
What is the maximum profit?
Answers
Answered by
Reiny
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 ? )
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 ? )
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