Question

A chain saw requires 4 hours of assembly and a wood chipper 6 hours. A maximum of 48 hours of assembly time is available. The profit is $150 on a chain saw and $220 on a chipper. How many of each should be assembled for maximum profit??

This is what I have... I don't know the process/ssteps to come to the answer:


Let c= chipper, let s= saw
obj: Profit=150s+220c
const: c>=0, s>=0
4s+6c<=48

Answer 1

is this being taught in calculus pr algebra II, there are a few ways to solve it. with calculus it is very easy

Answer 2

Yes, calculus

Answer 3

ok. so this is a standard optimization problem. start by isolating s or c

Answer 4

do you know what i mean by that?

Answer 5

u there?

Answer 6

Yes I get that... Isolating s or c

Answer 7

Would I isolate by kind of acting like i would solve for s or c

Answer 8

That's what you mean right

Answer 9

pick either s or c to isolate from the equation 4s +6c = 48

Answer 10

6c=48-4s

Answer 11

now divide both sides by 6...

Answer 12

ok I got it 2

Answer 13

C=8-.66s

Answer 14

how do i know the min and find the max profit?

Answer 15

now plug that in to your profit equation. now u can have everything in terms of s

Answer 16

now look for all times when the derivative of profit equation equals 0

Answer 17

you shouldnt have told me you were taking calculus. i started thinking about it like calculus. there is very easy way to do this. we now that 48 is divisible by 6 and it is also divisible by 4. so it is possible that you spend all your time on chainsaws or all your time on woodchippers. think about how much money you make per hour for each. for woodchipper its 220/6 =36.6 for chansaw its 150/4 = 37.5

Answer 18

you will make more per hour is spend all your time making chainsaws because 37.5 > 36.666

Answer 19

OOh arite. i gotcha. ... i missed the class when the professor was teaching this.. i was studying for another ex am =/

Answer 20

would i do the same process for this one as well:
deluxe coffee is to be mixed with regular coffee to make at least 50 pounds of a blended coffee. the mixture must contain at least 10 pounds of deluxe coffee. deluxe coffee cost $6 per pound and regular coffee $5 per pound. How many pounds of each kind of coffee should be used to minimize cost?

Answer 21

this one gets a little bit trickier because 6 doesnt go into 50 nicely

Answer 22

nevermind/ scratch that. its irelevant

Answer 23

lol ok.scratched

Answer 24

you want to make this as cheap as possible. but you need at least 10 pounds of deluxe. so use 10 pounds of deluxe. 10 x 6 = $60 dollars. for the other 40 pounds use regular. 40 x 5 = $200. total cost 60 + 200 = $260

Answer 25

it seems so easy when you know how to sort the info

Answer 26

so .. let x= deluxe, y=regular
\[x+y \ge 50 \]
constraints:\[x \ge60, y \ge40\] ?

Answer 27

i need to:
1. label all variables
2. show the objective function
3. list constraints
4. sketch the region and label all vertices defined by constraints
5. find the vertex that either min or maxed the fucntion

Answer 28

the constraints are x > or equal to 0 and y > or equal to 0

Answer 29

objective: 6x+5y=260

Answer 30

i isolated x and got x=4.3-3.3y .. is that my obj or did i have it correct above

Answer 31

i think whats mean by objective is to find the minimum cost

Answer 32

it hve to show an equation to minimize

Answer 33

so i guess that would be correct. i just think its strange to put the resultant minimum price in the objectice

Answer 34

idk my professor is a bit nutty... i must go to class now, thank so much for your help!