# linear programming

• May 25th 2008, 05:08 AM
maths4ever
linear programming
hello everyone...(Hi)

I have a quation here says
Bulldust Inc. blends silicon and nitrogen to produce two types of fertiliser. Fertiliser 1
must be at least 40% nitrogen and sells for \$20/kg. Fertiliser 2 must be at least 70%
silicon and sells for \$18/kg. Bulldust can purchase up to 800 kg of nitrogen at \$10/kg and
up to 1000 kg of silicon at \$8/kg. Assuming that all fertiliser produced can be sold,
formulate an LP to help Bulldust maximize profit.

thanks.
• May 25th 2008, 06:32 AM
CaptainBlack
Quote:

Originally Posted by maths4ever
hello everyone...(Hi)

I have a quation here says
Bulldust Inc. blends silicon and nitrogen to produce two types of fertiliser. Fertiliser 1
must be at least 40% nitrogen and sells for \$20/kg. Fertiliser 2 must be at least 70%
silicon and sells for \$18/kg. Bulldust can purchase up to 800 kg of nitrogen at \$10/kg and
up to 1000 kg of silicon at \$8/kg. Assuming that all fertiliser produced can be sold,
formulate an LP to help Bulldust maximize profit.

thanks.

Let $x_1$ and $x_2$ be the number of kg's of fertilizer 1 and 2 produced respectivley.

Let $y_1$ and $y_2$ be the proportion of N in f1 and f2 respectivly,

Then the cost is:

$c=x_1[10y_1+8(1-y_1)]+x_2[10y_2+8(1-y_2)]$

and the revenue is:

$r=20x_1+18x_2$

so the profit is:

$p=x_1[20-2y_1-8]+x_2[18-2y_2-8]=x_1[12-2y_1]+x_1[12-2y_2]$

This last is the objective (note its not linear unless $y_1$ and $y_2$ are fixed).

We have constaints:

$y_1 \ge 0.4$

$0 \le y_2 \le 0.3$

$y_1x_1+y_2x_2 \le 800$

$
(1-y_1)x_1+(1-y_2)x_2 \le 1000
$

$x_1 \ge 0$

$
x_2 \ge 0
$

Again note this is not a linear programming problem unless the $y$'s are fixed.

RonL
• May 25th 2008, 06:36 AM
masters
Quote:

Originally Posted by maths4ever
hello everyone...(Hi)

I have a quation here says
Bulldust Inc. blends silicon and nitrogen to produce two types of fertiliser. Fertiliser 1
must be at least 40% nitrogen and sells for \$20/kg. Fertiliser 2 must be at least 70%
silicon and sells for \$18/kg. Bulldust can purchase up to 800 kg of nitrogen at \$10/kg and
up to 1000 kg of silicon at \$8/kg. Assuming that all fertiliser produced can be sold,
formulate an LP to help Bulldust maximize profit.

thanks.

Show us one of your ways. Do you know what the answer is? Also, look here: http://www.mathhelpforum.com/math-he...lp-checks.html

This same question was posted by another member. Are you the same two persons?
• May 25th 2008, 06:43 AM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
Let $x_1$ and $x_2$ be the number of kg's of fertilizer 1 and 2 produced respectivley.

Let $y_1$ and $y_2$ be the proportion of N in f1 and f2 respectivly,

Then the cost is:

$c=x_1[10y_1+8(1-y_1)]+x_2[10y_2+8(1-y_2)]$

and the revenue is:

$r=20x_1+18x_2$

so the profit is:

$p=x_1[20-2y_1-8]+x_2[18-2y_2-8]=x_1[12-2y_1]+x_1[12-2y_2]$

This last is the objective (note its not linear unless $y_1$ and $y_2$ are fixed).

We have constaints:

$y_1 \ge 0.4$

$0 \le y_2 \le 0.3$

$y_1x_1+y_2x_2 \le 800$

$
(1-y_1)x_1+(1-y_2)x_2 \le 1000
$

$x_1 \ge 0$

$
x_2 \ge 0
$

Again note this is not a linear programming problem unless the $y$'s are fixed.

RonL

One thing you could do is to assume a number of mixtures for f1 and f2 and solve the resulting LPs and take as the full solution that of the mixture that gives tha largest maximum profit. (Of course it just might be that the 30/70 and 40/60 mixtures are the ones that maximise profits, but it will have to be demonstrated)

RonL
• May 25th 2008, 03:59 PM
deragon999
NO way
No we arent the same two people...we are just doing the same assignment