Expert: Josh - 1/9/2005

Question
The profit a coat manufacturer makes each day is modeled by the equation P(x)=-x^2+120x-2000, where P is the profit and x is the price for each coat sold.For what values of x does the company make a profit?
I don \'t understand this problem(how to do it) and hope you can help me.However, the answer key says that the answer is 20<x<100. Thanks.

Answer
Okay. The thing to appreciate is that any equation containing two variables, with one depending on the square of another describes a parabola in geometric terms.
In other words, things like y = a*x^2 + b*x + c can be visualize as a parabola in one form or another.

The given equation is clearly in this form [if you put y=P(x)]. In fact, because the coefficient for the leading coefficient (x^2) is negative (specifically, it is -1 here), it is an up-side-down parabola (looks like ^)
Note: If you are a little uncertain about this, refer to the little summary at the very end to refresh your memory.
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With this said, from 2D coordinate geometry, you know that a parabola of this form intercepts with the x-axis at (at most) two places. Consider the equation y=(x-q)(x-r). The idea is to find the values of "q" and "r", which are the solutions to this equation.
Refer to the diagram below.
 Y-AXIX
.....|
.....xxx
...x.|...x.
.-q--|----r--- X-AXIS
.x...|.....x
x....|......x

Since P(x) takes the place of Y in this problem, and it represents the cost, we make a profit if we manufacture X items and the value of the function is positive. (It means that we gain money, in return for what we produce).
If P(x) is negative, we lose money for the quantity of goods that we produce (usually, if we make too few or too many items).
If P(x)=0 exactly, we break-even.

Visually, we are interested in the portion between x=q and x=r, where the parabola (also called the operating curve) is positioned above the X-axis.

All-in-all, it amounts to solving the quadratic equation, using any of the following means.

a) by inspection,
b) factorization (or trial-and-error),
c) quadratic formula.

In any case, we get P(x)= -(x^2-120x+2000)=-(x-20)(x-100)
P(x)>0 in the interval 20<x<100.
You can just substitute an easy point into the equation to see this.
Say, pick x=0, P(x=0)=-(-20)(-100) is negative.
Pick x=220, you'll find P(x=220)=-(200)(120) is negative.

The way I select these test points is that I pick an easy number which is smaller than x=20, and another greater than x=100. The advantage with x=0,x=220 is that we can do this mentally without any difficulty.

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SUMMARY: Key things to remember about parabolas.

In general, it has the form y=a*x^2+bx+c or y=a(x-q)(x-r)  where x=q,x=r are the solutions to the equation.

Definitions:
If a is positive (a>0), it is upright (looks like U)
We describe this as a concave (or concave-up) parabola.

If a is negative (a<0), it is up-side-down (looks like ^)
We describe this as a convex (or concave-down) parabola.

Since x=q, x=r are the x-intercepts, their midpoint x=(q+r)/2 represents the axis of symmetry. Obviously, the shape of the parabola is symmetrical (a mirror image) about this line. The axis of symmetry must pass through the vertex (which is always a turning point in the context of a parabola).

If a>0, the vertex corresponds to a minimum value on the parabola.
If a<0, the vertex corresponds to the maximum value on the parabola.

To flip the parabola on its side, we exchange y with x and vice-versa, to obtain x=a*y^2+by+c or x=a(y-q)(y-r).
In this instance,
If a is positive (a>0), it looks like "("
[assuming the horizontal axis is labelled as X, and the value of x increases as we move toward the right.]
If a is negative (a<0), it looks like ")".