Question

90% of people can't solve this??? Maximizing income??

A manufacturer is producing two types of unites. Each unit Q costs 9$ for parts and $15 for labor and each uniot R costs $6 for parts and $20 for labor. The manufacturer's budget is $810 for parts and $1800 for labor. If the income per unit is $150 for Q and $175 for R, how many units of each should be manufactured to maximise income?

Answer 1

setup your constraints ....

Answer 2

How? @amistre64

Answer 3

this makes the ost sense to me

Q = 9p + 15l
R = 6p + 20l

parts: 9Q + 6R = 810
labor: 15Q + 20R = 1800

income: 150Q + 175R

Answer 4

not saying its right, but shouldnt these conditions hold?

Answer 5

Slowly starting to piece something together here :/

Answer 6

That's starting to make a lot more sense

Answer 7

I googled this question and got
"A manufacturer is producing 100 total units. Each unit Q costs $9 for parts and $15 for labor, and each unit R costs $6 and $20 for labor. The manufacturer’s budget is $810 for parts and $1800 for labor. If the income per unit $150 for Q and $175 for R, how many units of each should be manufactured to maximize income?"

Answer 8

0Q and 0R of course is the minimum income ... nothing made, nothing sold

lets just sell Q

9Q = 810 ; Q=90 <--- this limits out our budget sooo
15Q = 1500 ; Q=100

150(90) is our income for just selling Q within our budget

13500
------------------------------

maybe just R?

6R = 810 ; R=135
20R = 1500 ; R=75 <--- this limits out our budget sooo

175(75) = 13125
--------------------------------

spose we make 1R how many Qs can we make within budget?

9Q + 6 = 810 ; Q = 89 <---- limited to 89 to be in budget
15Q + 20= 1500 ; Q = 98
-------------------------------

income is: 150(89) + 175(1) = 13525

you see how this is working?

Answer 9

1800, not 1500 .... had some typos in mine

Answer 10

why Q+R = 100?

Answer 11

im getting 104 ... when i find the max values

Answer 12

I may have found a variation on the problem.

Answer 13

Q = 9p + 15l
R = 6p + 20l

parts: 9(9p + 15l) + 6(6p + 20l) = 810
labor: 15(9p + 15l) + 20(6p + 20l) = 1800

income: 150Q + 175R

p = 35/6
l = 1/2


Q = 9(35/6) + 15(1/2) = 60
R = 6(35/6) + 20(1/2) = 45

150(61) + 175(43) = 16875

Answer 14

ok, my notepad version of that has some sidereal calculation lol

60 and 45 is what i get as a max, and i tried to alter it to test

Answer 15

150(60) + 175(45) = 16875

150(61) + 175(43) = 16675 a decrease

Answer 16

Thanks so much

Answer 17

did you do something like this
http://prntscr.com/72qqwn

Answer 18

plug in the vertices

Answer 19

in a round about way yes

Answer 20

plugging in the corner points (0,0), (0,90), (60,45), and (90,0) into the objective function
income: 150x + 175y , it is maximized by (60,45)

Answer 21

So you can solve it two ways, looks like amistre did it by direct substitution, which makes sense since if you maximize the budget for each, you will get the most stuff to sell and therefore the most income.

As a linear programming problem:
Objective function: 150Q + 175R
Constraints:
9Q + 6R <= 810
15Q + 20R <=1800
Q>=0
R>=0

The last two constraints force us to look only in the first quadrant
and you will graph the lines
9Q + 6R = 810
15Q + 20R =1800 , find the vertices, plug them into the objective function