posted by on .

I really need help, this is my second post.

Information to solve problems
Q=12L+29L^2-1.1L^3

Q= # of cars produced/year
L=# of laborers used per year for production
Each car sells for \$125.
\$7000 per laborer per year is cost for labor and other consumable materials.

Questions:

1. What is the max and min number of laborers possible for the production of cars? This is the domain of what function?
I really don't get the domain of what function part.

2. How many laborers would be needed to maximize the production of cars?

3. How many laborers would be needed to maximize the revenue for producing cars? What is the maximum revenue?

4. Is there a range of values of production for the cars that would be profitable? IF so, what is the range?

1.
The domain of a function f(x) is the range of values of x which is admitted by the function f.
For example, the domain of
f(x)=x²+2x-4
is (-∞,+∞), or all real numbers.
On the other hand, the domain of
f(x)=ln(x) is (0,∞), since the function ln(x) cannot accept non-positive values of x.
In this case, since we cannot hire a negative person, so the domain starts at ______ and goes up to _______.

The domain applies to the function _______=12L+29L^2-1.1L^3.

2.
For the maximum production, you will need to find dQ/dL and equate it to zero to find the maximum (or minimum).

3. Revenue R(L) is the selling price minus the cost.
Selling price = \$125*Q(L)
cost = \$7000L
So R(L)=125Q(L)-7000L
Again, find L for maximum R(L) by equating dR/dL=0.

4. Profitable means R(L)>0.
Find the interval on which R(L)>0, i.e. find the lower and upper limits of L between which R(L)>0.
Note that L must be an integer.

1. So this is the answer I got 12L+29L^2-1.1L^3=0
L(12+29L-1.1L^2)=0
L=0, 26.77, -0.41
The minimum number of laborers would be 0.
The maximum number of laborers would be 26.77.
THe range would then be (0,26.77) because when I substitute a number >26.77 into the above equation it becomes a negative number. Am I on the right track or still way off?

2.Q'=12+58L-3.3L^2
12+58L-3.3L^2=0
L=17.78, -0.20
Q'(max)=17.78
To maximize the production they need 17.78 laborers

3. TO figure out how many laborers would be needed to maximize the revenue for producing cars?

How would I set that up?

4.Profit = price*production-7000L
L(-137.5L^2+3625L-5500)=0
L=0, 1.56, 24.9
The range would be 1.56 - 24.9. (I'm not sure if this is right).

How would I justify this answer with a graph? SHow the graphs of the 1st and 2nd derivatives?

(!!!)L>=0 integer.
1)Q=0 if L=-0.41 or L=0 or L=26.77
Q>=0 if 0<=L<=26
2)Q'=12+58L-3.3L^2=0 if L=17.78
Q(17)=3180.7 and Q(18)=3196.8, therefore L=18
3)The revenue R=(12L+29L^2-1.1L^3)x125-7000L
R=-5500L+3625L^2-137.5L^3
R'=-5500+7250L-412.5L^2=0 if L=16.75
R(16)=276800 and R(17)=278587.5,
therefore L=17
4)R=0 if L=0 or L=1.6 or L=24.7
R>=0 if 2<=L<=24