# Thread: Quadratic optimisation

1. ## Quadratic optimisation

A manufacturer finds that the profit P from assembling x bicycles per day is given by

P(x) = -x^2 + 50x - 200

What is the maximum profit?

I figured to find the max profit i should find the turning point of of the parabola which illustrates this equation.

What I got was (25, 1675)

So I figured that it would be $1675 HOWEVER, the answer says 425 dollars! Could you tell me how to do this? 2. Originally Posted by juliak A manufacturer finds that the profit P from assembling x bicycles per day is given by P(x) = -x^2 + 50x - 200 What is the maximum profit? I figured to find the max profit i should find the turning point of of the parabola which illustrates this equation. What I got was (25, 1675) So I figured that it would be$1675

HOWEVER, the answer says 425 dollars!

Could you tell me how to do this?

i think
Maximum profit P(x)= -x^2+50x-200 at maxima x=25 will be
P(25) = -(25)^2+50*25-200
= -625+1250-200
= 425

3. Why is the negative sign not squared to make term a positive?

4. Originally Posted by juliak
Why is the negative sign not squared to make term a positive?
-x^2 does not mean (-x)^2. It means -(x^2).

5. Originally Posted by ramiee2010
i think
Maximum profit P(x)= -x^2+50x-200 at maxima x=25 will be
P(25) = -(25)^2+50*25-200
= -625+1250-200
= 425
Do you think you might tell us why you think that.

CB

6. Mr Fantastic just want to be helpful, things that might be clear to us, but not to newbies.

7. P(x) = y = -x^2 + 50x - 200,

x^2 - 50x +200 = -y

completing the square,

x^2 - 50x +(-50/2)^2 = -y +(-50/2)^2 - 200

(x - 25)^2 = -(y - 425)

(x,y) --> (25, 425), the value of x and y are the maximum that can be obtained, its the apex of the inverted parabola

x = units,

y = P(x) = \$425.00, profit in dollars

the graph below confirms the answer.