1. We were given this as the function: Q=12L+29L2-1.1L3 we know that L is laborers and Q is the production of Bangalores. First I’m going to factor one L from the function.
Q=L(12+29L-1.1L2) from here I can’t factor inside the parenthesis so I will use the quadratic formula. From here I get L=-0.407495 and L=26.77131. We can’t have a negative amount of Laborers so I would say that the maximum number of employees is 26. I can see the L from the factored equation can be set to zero also so the minimum number of employees would be zero. Because we are solving for L I guess you could say it’s the domain of the labor function.
2. To find the number of laborers needed to maximize the production of bangalores I need to find the critical point by first taking the first derivative. Q=12L+29L2-1.1L3 Q’=12+58L-3.3L2 .now that I have this new equation I can set it equal to zero so I can find the critical points.
12+58L-3.3L2=0 I can’t factor so I will need to use the quadratic formula again.
= = L= -.204517 and L=17.780274 in the first problem we have the domain as (0,26) -.204517 isn’t in the domain so the maximum number of laborers is about 18.
3. I know that the equation for revenue is this:Revenue = unit cost * quantitySo I know that the unit cost of bangalores, because we want the revenue for producing bangalores, is $125. Q is the numbers of bangalores produced per year so I will need to find the function Q. In number 2 we found that about 18 Laborers maximize the production of bangalores so I would think that it would also maximize revenue as well. I’m going to plug 18 into the function. Q=12(18)+29(18)2-1.1(18)3 = 3196.8 we know that this is the quantity.
Maximum Revenue= 125*3196.8= $399600.00
4.I found this equation for profit P = R - C this is where the profit equal the revenue –C which is the $7000 per laborer year. R=125*Q and C= 7000L (L for per laborer year) From here all that is left to do it plug in the function Q. P= 125(12L+29L2-1.1L3)-7000L
=1500L+3625L2-137.5 L3–7000L= -5500L+3625L2-137.5L3 I can see that I can factor one L out of the equation. P=L(-5500+3625L-137.5L2) We can set L to zero so we know that L=0. The part in the parenthesis I can’t factor so I will be using the quadratic formula again.
= L= 1.616338 and L=24.74298 The range is between 2 and 28.
A graph of P= -5500L+36256L2-137.5L3:
5. My answer from number 3 would not give the maximum profit for producing bangalores because the revenue from that equation was the maximum revenue which was $399600.00. For number four we needed the revenue to have L’s because to find the range we need to be able to find the values for L. In number 3 L= 18 where number 4 it was just the equation Q without being solved. We wouldn’t be able to get a range from 399600.00-7000.
6.This problem is very similar to number 2. So to start I will be taking the derivative of
P=-5500L+3625L2-137.5L3(because we are looking for the maximum profit this time)
which is P’=-5500+7250L-412.5L2we can now set P’ equal to zero and find the critical numbers. P’=-5500+7250L-412.5L2=0 I can’t factor this so I will be using the quadratic equation again.
= L= .794539 and L=16.7812119 Now I need to find the second derivative of the function P so I can find where the graph will be at a maximum and where it will be at a minimum. P’’= 7250 - 825L I can now plug in my L values into the function P’’ P’’(.794539)=7250-825(.794539)=6594.505325
P’’(16.7812119)=7250-825(16.7812119)=-6594.499818. because P’’(17) is equal to something to something smaller than zero L= 17 is the maximum profit.
Graph of P’:
Graph P’’:
7.