Posted by Janet on Thursday, April 21, 2011 at 11:15pm.
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?
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