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February 14, 2016
Posted by **Janet** on Thursday, April 21, 2011 at 11:15pm.

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?

- Calculus--Please help. -
**MathMate**, Thursday, April 21, 2011 at 11:32pm1.

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.

- Calculus--Please help. -
**Janet**, Friday, April 22, 2011 at 12:27am1. 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?

- Calculus--Please help. -
**Mgraph**, Friday, April 22, 2011 at 12:55am(!!!)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

- Calculus--Please help. -
**Janet**, Friday, April 22, 2011 at 11:28pmCarry on with the problem.. To find maximum profit, not revenue, would I take the derivative of the profit function, which I have as

Profit=125*(12L+29L^2-1.1^3)-7000L. If I take the derivative of this(which I'm not sure I did right) my L=16.78 or .79. If I plug it back into the original profit function my max comes out to 278746.76. Does this look right?