1. ## compute income

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2. 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

3. 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?

4. 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