1 Final Calc question! Profit maximization problem

[phatj]

Still studying for that test... came across this problem and it is annoying me so much!!

Jane, a farmer, can raise hogs, cattle, and sheep.

She has room for 80 sheep, or 120 hogs, or 60 cattle.

so 8s=12h=6c

State law requires that Jane raise as many hogs as cattle and sheep combined.

so h = c + s

She estimates profit of $10 per sheep, $8 per hog, and $20 per cow. How can she maximize her profit?

Here's what I've done, but deriving this doesn't make much sense...

8s=12h=6c
h= c + s

Profit (c,h,s) = 10s + 8h + 20c
Profit (c,s) = 10s + 8(c+s) + 20c

Deriving that does not give me anything useful, right? So where am I going wrong? Thanks!

 

[rudder]

Maximize his profit by having him use some field space to grow some marijuana.

 

[MacBaine]

She needs to inflate her records and manipulate her stock prices, then when she is indicted, leave with a multi-million dollar pension.

 

[phatj]

.....

 

[phatj]

anyone?

 

[ATLien247]

I ain't gonna solve it for you, but you might want to use partial derivatiives and the Lagrange multiplier.

 

[phatj]

I'm not asking for someone to solve it for me, just help me out -- this problem shouldn't utilize the Lagrange multiplier, the problem comes before that topic is covered in the book. Using partial derivatives gets me nowhere using the P(c,s) equation I have.

 

[Stunt]

can you take ur profit equation and sub in the equivalancy formula so you have only one variable...derive...set to zero, find the number of say cattle?

 

[SurgicalShark]

Here is your answer:

maximize 10c + 8h + 20c
subject to

s <= 80
h <= 120
c <= 60
h - c- s = 0

h >= 0
c >= 0
s >= 0


LP OPTIMUM FOUND AT STEP 3

OBJECTIVE FUNCTION VALUE

1) 2760.000

VARIABLE VALUE REDUCED COST
C 60.000000 0.000000
H 120.000000 0.000000
S 60.000000 0.000000


ROW SLACK OR SURPLUS DUAL PRICES
2) 20.000000 0.000000
3) 0.000000 8.000000
4) 0.000000 30.000000
5) 0.000000 0.000000
6) 120.000000 0.000000
7) 60.000000 0.000000
8) 60.000000 0.000000

NO. ITERATIONS= 3

It means that you need to have 60 cattles, 120 hogs and 60 sheeps for total profit of $2760

 

[phatj]

shark, what program did u use?

 

[phatj]

and, that doesn't really show me HOW to do it.....

 

[SurgicalShark]

It is my software to solve the LP models.

 

[ActuaryTm]

Office hours exist for a purpose.

 

[Jassi]

Originally posted by: SurgicalShark

It means that you need to have 60 cattles, 120 hogs and 60 sheeps for total profit of $2760

She has room for 80 sheep, or 120 hogs, or 60 cattle.

Notice the OR 😉 You assumed AND, so your answer is wrong. That statement is confusing to me. Maybe assume that you have 240 units of area, hog takes 2 each, sheep takes 3 each and cattle takes 4 each. Now, you have a constant, the total area must equal 240. That gives

(sheep*3) + (hog*2) + (cattle*4) = 240

Condition - hogs = sheep + cattle; substitute into hog

(sheep*5) + (cattle*6) = 240 (1)

Now, set up a function for profit and you might have a system to solve the problem.

Profit = (18*sheep) + (28*cattle) (2)

Solve for cattle in (1), plug in to (2) and use the derivative method to find max profit.

 

[Jassi]

Hey OP, was I right?