# compute income

• Aug 25th 2007, 03:49 PM
Micka1
compute income
132
• Aug 25th 2007, 05:06 PM
TKHunny
Price * Number Sold = Total Income.

x(y) = Price based on number sold = 400 - 0.3y

y = Number sold

y*x(y) = Total Income = z(y) = y*(400-0.3y) = 400y - 0.3y^2
• Sep 1st 2007, 10:24 PM
TKHunny
z(y)=400y-0.3y^2
v(y)=4000+6y-.001y^2

$Profit(y) = z(y)-v(y) = 4000 + 394y - 0.299y^{2}$

Since you are to use Differential Calculus,

$\frac{dProfit}{dy} = -0.598y + 394$

Where is that zero?

-0.598y + 394 = 0 and Solve for y.

You are beginning to worry me. It seems as though you are a little too eager to give up on a problem if it does not immediately make sense. Maybe it would help to know why you are in this class. Have you had success in the prerequisite mathematics courses?
• Sep 5th 2007, 06:45 AM
topsquark
Quote:

Originally Posted by Micka1
4000 + 394y - 0.299y^2
Using the quadratic formula gives you
y=1327.80, y=-10.07
To find which is the max,
-0.598y + 394
substitute y,
y = 1327.80 = -400.02 y = -10.07= 400.02
y = 1327.80 = -400.02 is a negative
therefore 1327.80 is the maximum.
Any thoughts? Have i got this right?

What??

You solve
$\frac{d}{dy}(4000 + 394y - 0.299y^2) = -0.598y + 394 = 0$
to find the critical values of y, just as TKHunny showed you. Then you look at each of these y values in the original equation to see which is the minimum or maximum. (Or you can use the second derivative test.)

-Dan