Linear Programming: Optimization

buddha1119 · #1 $Old$ October 19th, 2009, 01:52 PM

Hello,
I would like to learn how to solve this problem without Excel Solver. Can someone please show me the steps on how to simultaneously solve the equations from this problem? Thanks.

A farmer has 900 acres of land. The farmer is going to plant each acre with corn, soybean, or wheat.

Each acre planted with wheat yields 2k profit.
Each acre with soybean yields 2.5k profit.
Each acre with wheat yields 3k profit.

The farmer has 100 workers.
The farme has 150 tons of fertilizer.

Per acre
Corn requires: .1 worker, .2 tons of fertilizer
Soubean requires: .3 worker, .1 tons of fertilizer
Wheat requires: .2 worker, .4 tons of fertilizer

objective:
2000Corn + 2500Soy + 3000Wheat = Max Profit

Equations.
XCorn + YSoy + Zwheat <= 900
.1Corn + .3Soy + .2wheat <=100
.2Corn + .3Soy +.4wheat <= 150

snaes · #2 $Old$ October 19th, 2009, 08:50 PM

Quote:

Originally Posted by buddha1119 $View Post$

Each acre planted with wheat yields 2k profit.
Each acre with soybean yields 2.5k profit.
Each acre with wheat yields 3k profit.

Top one should be corn (I think)...I'll try and figure this out too.

I believe the best way would be lagrange multipliers (requireing calculus)...

EDIT: Do you know the answer?

snaes · #3 $Old$ October 19th, 2009, 09:26 PM

EDIT: THIS IS WITH ONE CONSTRAINT, THE PROCESS IS SIMILAIR WITH TWO CONSTRAINTS, THE ONLY DIFFERENCE IS YOU GET 5 EQUATIONS WITH 5 VARIABLES, NOT 4 AND 4. THIS ANSWER IS NOT CORRECT, BUT THE PROCESS CORRECT.

Maximize: F
$2000c+2500s+3000w$

Constraints:
$.1c+.3s+.2w=100$
$.2c+.3s+.4w=150$
I set them = not less than or equal to because we truely want to maximize using on contraint then try it using the other.

$\nabla F = <2000c,2500s,3000w>$
set this equal to the gradient of a constraint:
$<2000c,2500s,3000w>=\lambda<.1,.3,.2>$

Get all the equations:
$2000c=\lambda.03$
$2500s=\lambda.09$
$3000w=\lambda.04$

Solve for lambda ( $\lambda$ ) and set top 2 equal:
$2000c/.03=2500s/.09$
Same thing for bottom 2 equations
$2500s/.09=3000w/.04$

after simplifying:
$2.4c=s$
$10s=27w$

adjust ratios to get all of them "equal"
$24c=10s=27w$

Solve for "c" in terms of s and then in terms of w serpately.
$2.4c=s$
$8/9c=w$

plug in first constraint:
[math].1c+.3(2.4)c+.2(8/9)c=100
$c=100.222$
$s=240.535$
$w=88\frac{8}{9}$

put these back into the equation you want to maximize (profit)
you get about: $ $1069042.5$

Wilmer · #4 $Old$ October 19th, 2009, 10:17 PM

Perhaps this'll help you: Linear programming - Wikipedia, the free encyclopedia

snaes · #5 $Old$ October 20th, 2009, 12:01 PM

Here is how to set it up. It would be the exact same process as before only with the added constraint.

[math]<2000c,2500s,3000w>=\lambda<.1,.3,.2> + \mu<.2,.3,.4>

Now you have 5 equations and 5 variables. Now all you need to do is solve. Generally solving for $\lambda$ and setting things equal helps simplify.