Question

The price of a certain monthly magazine is x dollars per copy. The total profit from the sales of the magazine is P dollars. It is given that P=Y+Z, where Y varies directly as x and Z varies directly as the square of x. When x is 20, P is 80000; when x is 35, P is 87500.
(a) Express P in terms of x.
==>P=6000x-100x^(2)
(b) Find the profit when x=15.
==>67500
(c) Using the method of completing the square, express P in the form a-b(x-c)^(2), where a,b and c are constants. Find the values of a,b and c.
==>???
(d) Hence, or otherwise, find the value of x when P is maximum.
==>???

Answer 1

(a) and (b) are good.
for part (c)

you start with
\[ -100 x^2 +6000 x = P \]

for how to complete the square see
http://www.khanacademy.org/math/algebra/quadratics/completing_the_square/v/completing-the-square

Answer 2

but in the meantime,

we want the x^2 term to have a coefficient of 1
(because this makes "completing the square" easier)
one way is divide both sides of the equation (all terms) by -100
\[ x^2 -60x = \frac{P}{-100} \]

the next step is divide the number in front of the x by 2: -60/2= -30
(see the video for why we do this)
remember that -30

next square the -30 to get 900
we add this 900 to both sides of the equation
\[ x^2 -60x +900 = \frac{P}{-100} +900\]

now it may not be obvious, but the stuff on the left side of the = is the same
as (x-30)^2 (that is why we want to remember the -30)
As a check, use FOIL: (x-30)*(x-30)= x^2 -30x -30x +900= x^2-60x+900

so we now have
\[ (x-30)^2 = \frac{P}{-100} +900\]

Answer 3

They want
express P in the form a-b(x-c)^(2), where a,b and c are constants. Find the values of a,b and c.

\[ (x-30)^2 = \frac{P}{-100} +900 \]
so let's solve for P. add -900 to both sides
\[ \frac{P}{-100} +900 -900 = -900 + (x-30)^2 \]
or
\[ \frac{P}{-100} = -900 + (x-30)^2 \]
multiply both sides by -100:
\[ P= 90000 - 100 (x-30)^2 \]
that is the from they want, where a is 90,000 b is -100 and c is 30

Answer 4

**b is +100
because they use the form
a-b(x-c)^(2)
and we have
90000 - 100(x -30)^2

Answer 5

to do (d), look at the equation
P= 90000 - 100 (x-30)^2

notice it is 90,000 MINUS something.
for P to be as large as possible we want to subtract as little as possible.

the term 100*(x-30)^2 is either 0 or positive (because even if x-30 is a minus, after squaring it, it will be positive. The smallest it can be is zero.
when it is zero, P will be the largest it can be.

what x value makes 100*(x-30)^2 zero ?