# Optimization

• Nov 9th 2009, 12:53 PM
Dogod11
Optimization
Hello, at first had this problem:

A mountain lake in national park is stocked each spring with two species of fish, S1 and S2. The average weight of the fish stocked is 4 pounds for S1 and 2 pounds for S2. Two foods, F1 and F2, are available in the lake. The average requeriment of a fish of species S1 is 1 unit of F1 and 3 units of F2 each day. The corresponding requeriment of S2 is 2 units of F1 and 1 unit of F2.

a). Suppose that 1000 units of F1 and 1800 units of F2 are available daily in example previous. How should the lake be stocked to maximize the weight of fish supported by the lake?

Well, I have already found the solution, graphical method using the search

Maximize $4x_1 + 2x_2$

subject to restrictions

$x_1 + 2x_2 \leq{1000}$

$3x_1 + x_2 \leq{1800}$

And the solution is $x_1 = 520, x_2 = 240$

The question I have is to resolve this part:

In problem previous, how should the lake stocked to maximize the total number of fish supported by the lake?

• Nov 11th 2009, 02:21 AM
BobP
THe second part can be answered graphically as well, but here is an algebraic solution.
We have,
$x_1+2x_2\leq1000$ and $3x_1+x_2\leq1800.$
Add twice the first inequality to the second and you have
$5x_1+5x_2\leq3800.$
Divide by 5 and you get
$x_1+x_2\leq760.$
• Nov 11th 2009, 04:46 AM
Dogod11
But keep in mind that this problem comes third restriction??
$
4x_1 + 2x_2 = 2560.$
• Nov 11th 2009, 05:57 AM
BobP
That isn't a problem is it ?
$x_{1}=520, \quad x_{2}=240,$ is at the limit of $x_{1}+x_{2}\leq760.$
What this is saying is that the two alternative requirements, (1) maximize the weight and (2) maximize the number, lead to the same result.
Am I missing something ?
• Nov 11th 2009, 06:17 AM
Dogod11
Hello,

I think it should be as you say

but I was posing like this: Since the first part of the exercise was found the maximum weight supported by the lake

$4(520) + 2(240) = 2560$

so now the problem is

maximize $P = x_1 + x_2$

subject to restrictions

$4x_1 + 2x_2 = 2560$

$x_1+2x_2\leq1000$

$3x_1+x_2\leq1800.$

this is wrong, or gives the same solution?

Thank you very much ...
• Nov 11th 2009, 06:27 AM
Dogod11
The result I get is $P = 760,$

$x = 520, y = 240$

well, anyway it does not matter jeje

Greetings
• Nov 11th 2009, 06:42 AM
CaptainBlack
Quote:

Originally Posted by Dogod11
Hello, at first had this problem:

A mountain lake in national park is stocked each spring with two species of fish, S1 and S2. The average weight of the fish stocked is 4 pounds for S1 and 2 pounds for S2. Two foods, F1 and F2, are available in the lake. The average requeriment of a fish of species S1 is 1 unit of F1 and 3 units of F2 each day. The corresponding requeriment of S2 is 2 units of F1 and 1 unit of F2.

a). Suppose that 1000 units of F1 and 1800 units of F2 are available daily in example previous. How should the lake be stocked to maximize the weight of fish supported by the lake?

Well, I have already found the solution, graphical method using the search

Maximize $4x_1 + 2x_2$

subject to restrictions

$x_1 + 2x_2 \leq{1000}$

$3x_1 + x_2 \leq{1800}$

And the solution is $x_1 = 520, x_2 = 240$

The question I have is to resolve this part:

In problem previous, how should the lake stocked to maximize the total number of fish supported by the lake?.

You now have to maximise $x_1+x_2$ subject to the same constraints:

$x_1 + 2x_2 \leq{1000}$

$3x_1 + x_2 \leq{1800}$

CB
• Nov 11th 2009, 06:45 AM
CaptainBlack
Quote:

Originally Posted by Dogod11
Hello,

I think it should be as you say

but I was posing like this: Since the first part of the exercise was found the maximum weight supported by the lake

$4(520) + 2(240) = 2560$

so now the problem is

maximize $P = x_1 + x_2$

Yes

Quote:

subject to restrictions

$4x_1 + 2x_2 = 2560$

No, you cannot generaly get the maximum mass of fish with the maximum number.

Quote:

$x_1+2x_2\leq1000$

$3x_1+x_2\leq1800$

this is wrong, or gives the same solution?

Thank you very much ...
• Nov 11th 2009, 06:46 AM
Dogod11
anyway gives the same result: $x = 520, y = 240.$

Thank you very much....