Expert: Steve Holleran - 7/17/2007

Question
okay so i have a couple questions my teacher gave us a take home summative and we're allowed to use any means to find the answers...I'm wondering if you can help.  I did them all. Here's the ones I'm not sure i got the right answers to;

show that for any real number m, the quadratic equation
mx^2 +(3m+2)x + (2m+3)=0

will always have two distinct roots..

solve for x in the equation (I got three different answers)

1/3 - 2/3x-1 = x+1 / -2x+3
finally,
the demand equation for a product is p = - 3x +25 where p is the price in dollars and X is the number of items sold in thousands. the cost function is C(X)= 7x +15
a)find the corresponding revenue function
b)find the corresponding profit function
c) complete the square to show the maximum profit
d) find the "break even" quantities
 
any help will be greatly appreciated

thanks again
meghan  

Answer
Hello Meghan,

On the first one:

For the nature of the roots of a quadratic, all you have to look at is the value of the discriminant (the stuff under the square root in the quadratic formula).  So, if we evaluate the  b^2 - 4ac part of the formula here, we have

a = m     b = (3x+2)   c = (2m+3)

and

b^2 - 4ac = (3m+2)^2 - 4(m)(2m+3)

         = 9m^2 + 12m + 4 - 8m^2 - 12m  = m^2 + 4

Since this is always positive, the discriminant is always positive, and this means there are two distinct roots.


For the second part:

Use the quadratic formula with the a, b, c values above.
We already know the part under the square root.

  x = [-(3m+2) +/- sqrt(m^2 + 4)] / (2m)

So you have two solutions:


  [-(3m+2) + sqrt(m^2 = 4)] / (2m)

and [-(3m+2) - sqrt(m^2 + 4)] / (2m)



On this third one, I hope I interpret what you have written correctly.  I'm assuming the price equation is per item.

So, for part (a),
Revenue = (# items sold) * (price per item)

     R = 1000x * (-3x + 25)

     R = -3000x^2 + 25000x

B) Profit = (# items) * (price - cost)

       P = 1000x * [(-3x + 25) - (7x + 15)]

         = 1000x * [ -10x + 10]

       P = -10,000x^2 + 10,000x

C) For this, we have to factor -10,000 out of the right side first:

       P = -10,000 * [x^2 - x]

Now to complete the square, we take half the linear coefficient and square it:   1 divided by 2 = 1/2,
(1/2)^2 = 1/4.  Now add 1/4 on the right, in the brackets .

       P = -10,000 * [x^2 - x + 1/4]

Okay, now we've got to balance this out.  We just added a "1/4" in the brackets, but notice that the stuff in the brackets is being multiplied by -10,000, so we really added
-10,000 * 1/4 = -2,500 to the right side.  So to balance we need  to add +2,500:

         P = -10,000 * [x^2 - x + 1/4] + 2500

Then      P = -10,000 * (x - 1/2)^2 + 2500

So then the maximum occurs when x = 1/2 (that is, 1/2 of a thousand units, or 500 units), and the max profit is $2500.

For the break even point, set P = 0 in the original equation for profit:

           0 = -10,000x^2 + 10,000x

   10,000x^2 = 10,000x

         x^2 = x   so x^2 -x = 0

                      x(x-1) = 0   and x = 0 or 1

Which means the break even point is at x = 1 which is 1000 units.

I hope this helps you out, Meghan

Steve Holleran