Expert: Paul Klarreich - 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
Questioner:   meghan
Category:  Advanced Math
Private:  No
 
Subject:  math help
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
.....................................
Hi, Meghan,

So you're trying to understand equations, both the simple and quadratical.  

1. This:  mx^2 +(3m+2)x + (2m+3)=0
is a quadratic equation.  The nature of the roots is given by the 'Discriminant' part of the quadratic formula:

D =  b^2 - 4ac,  normally found under a radical sign.

If  D = 0, then the roots are identical.  So you have to show that D /= 0, so the roots are not identical.

a = m,  b = 3m + 2,   c = 2m + 3

D = (3m + 2)^2 - 4(m)(2m + 3)

D = 9m^2 + 12m + 4 -  8m^2 - 12m

D = m^2 + 4

That is clearly positive, since:

m^2 >= 0  for all real values of m.
4   >  0
-------------
m^2 + 4 > 0

So your D is always positive, never zero, and the roots are real and distinct.

.......................................
2. solve for x in the equation  1/3 - 2/3x-1 = x+1 / -2x+3

I assume you mean this equation:  (If you misparenthesized, then I don't have the right problem here.)

1        2      x + 1
--- -  ------ = -------
3     3x - 1   -2x + 3

Combine terms on the left, using  3(3x - 1) as the LCD.
3x - 1 - 6     x + 1
----------  = -------  
3(3x - 1)     -2x + 3

3x - 7      x + 1
-------  = -------  
9x - 3     -2x + 3

Cross-multiply:

(3x - 7)(-2x + 3) = (9x - 3)(x + 1)

- 6x^2 + 14x + 9x - 21  = 9x^2 - 3x + 9x - 3

- 6x^2 + 23x - 21  = 9x^2 - 6x - 3

Get all the terms to the right side:

0 = 15x^2 - 29x + 18

Now this is a quadratic.  Does it factor?  Look at that D again:

D = 29^2 - 4(15)(18)

D = 841 - 4(270) = 841 - 1080 = - 239

A negative discriminant means imaginary roots.  I suspect there is some error in transcribing the example, so I'm going to stop here.

.......................................
3 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

I don't generally handle 'business' problems, but this isn't really one of them.  I assume that:

Revenue = items sold * price, and
Profit = Revenue - total cost.

[Even us academic types know that.]

Revenue(x) = 1000x(-3x + 25) = -3000x^2 + 25000x

and I think that the cost function represents the unit cost -- the cost to make ONE umbrella.  (Is that what you are selling?)

In that case the total cost is:

Total cost(x) = 1000x C(x) = 1000x(7x + 15) = 7000x^2 + 15000x

Profit(x) = -3000x^2 + 25000x - (7000x^2 + 15000x)

Profit(x) = -3000x^2 + 25000x - 7000x^2 - 15000x

Profit(x) = -10000x^2 + 10000x

Profit(x) = -10000(x^2 - x)

Now we can complete the square.

Profit(x) = -10000(x^2 - x + 1/4 - 1/4)

Profit(x) = -10000(x^2 - x + 1/4)  + 2500

Profit(x) = -10000(x - 1/2)^2  + 2500

OK, then:  Your maximum profit comes when that HUGE negative thing at the left becomes zero.  That occurs when x = 1/2, or 500 radios,,... I mean, umbrellas.

And the profit will be  2500.