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?
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 ? )
To find the number of items that should be produced to maximize the profit and the maximum profit itself, we need to analyze the given profit equation.
The profit equation is P = -40x^2 + 1000x - 5250, where P is the profit and x is the number of items produced.
To find the number of items that yield maximum profit, we need to find the vertex of the quadratic equation. The x-coordinate of the vertex can be found using the formula x = -b / (2a), where a, b, and c are the coefficients of the quadratic equation in the form ax^2 + bx + c.
In this case, a = -40 and b = 1000. Plugging these values into the formula, we have x = -1000 / (2 * -40) = 12.5.
So, the number of items that should be produced to maximize the profit is 12.5 (typically rounded to the nearest whole number), which means 13 items should be produced.
To find the maximum profit, we substitute the value of x = 13 back into the profit equation:
P = -40(13)^2 + 1000(13) - 5250
= -40(169) + 13000 - 5250
= -6760 + 13000 - 5250
= 9970.
Therefore, the maximum profit is $9970.