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
Carry 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?
To solve these problems, we need to analyze the given information and understand the concepts related to functions, optimization, and profitability. Let's address each question step by step:
1. The function provided in the problem, Q = 12L + 29L^2 - 1.1L^3, relates the number of laborers used per year (L) to the number of cars produced per year (Q). The domain of a function refers to the set of all possible input values for the independent variable, in this case, the number of laborers (L). However, the given function does not have any restrictions on L, so the domain is all real numbers.
2. To determine how many laborers are needed to maximize the production of cars, we need to find the value of L that corresponds to the maximum value of Q. This can be done by finding the derivative of the function with respect to L and setting it equal to zero. Then solve this equation for L to obtain the number of laborers that maximize the production.
3. To find the number of laborers needed to maximize the revenue, we need to multiply the number of cars produced (Q) by the selling price of each car ($125). The revenue function is given by R = 125Q = 125(12L + 29L^2 - 1.1L^3). Similar to the previous step, we find the derivative of the revenue function with respect to L, set it equal to zero, and solve for L to determine the number of laborers that maximize the revenue. The maximum revenue can be calculated by substituting the corresponding value of L into the revenue function.
4. Profitability depends on various factors, such as costs and revenues. To determine the range of production values that are profitable, we need to consider the costs associated with labor and materials. In the given problem, the cost per laborer per year is $7000. Using this information, we can calculate the total cost as C = 7000L. By subtracting the cost from the revenue, we can determine the profit, which is given by P = R - C. The production range that is profitable corresponds to the values of Q (number of cars) for which P (profit) is positive.