# Basic system of linear inequalities in two variables question

#### basicmathguy1993

I just wanted to make sure that with this particular problem, the designated independent and dependent variable doesn't matter.

Problem:
Farmer McGregor plants oats and wheat on his farm. For conservation purposes, he plants at least twice as many acres of wheat as oats. He can handle up to a total of 540 acres of planting. What combinations of plantings can he consider?

My solution was:
x = acres of wheat
y = acres of oats

Therefore,
y <= -x + 540
and
y<= (1/2)x

Obviously, Constraints x >=0, y >= 0

This creates a shaded inequality along the x-axis as in the first attached graph. The solutions manual designates x = acres of oats, y = acres of wheat and gets a shaded inequality graph along the y-axis. So basically, it the independent and dependent variables don't matter in a problem like this?

Thanks for the help,

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#### romsek

MHF Helper
The way I'd do this is to assume that he's going to use all 540 acres.

let $\alpha = \dfrac {wheat}{oats} \geq 2$

$540 = wheat + oats = (1+\alpha)oats$

Next we can plot land utilization as a function of $\alpha$

You can of course choose to use less than 540 acres. Choosing $N$ acres will result in

$N = (1+\alpha)oats$

$wheat = \alpha\cdot oats$

#### JeffM

$Let\ z = total\ acres\ planted.$

$Let\ x = acres\ planted\ in\ wheat.$

$Let\ y = acres\ planted\ in \ oats.$

$z = f(x,\ y) = x + y.$

The dependent variable is z, and x and y are both independent variables, and each variable is subject to at least one constraint, namely:

$x \ge 0;$

$y \ge 0;$

$0 \le z;$

$z \le 540;\ and$

$x \ge 2y.$

Whether you define acres of wheat as x or y is completely arbitrary. In fact you could label acres of wheat as w, and acres of oats as o, and your perplexity would immediately disappear. The assignment of letters to variables is purely arbitrary.

EDIT: Romsek has shown a different way to set this up. I prefer my way because if this problem was defining the feasible region in a linear programming problem, the answer would lie at a corner of the feasible region, in this case the green triangle.

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#### basicmathguy1993

@romsek & JeffM: Thank you very much for the response. I have to apologize upfront because I neglected to state the textbook instructions: "Write a system of linear inequalities in two variables to represent each situation and graph. Determine one possible solution from the graph." The main concern I have here is how you choose the independent and dependent variable. It seems to me that it is arbitrary in a problem like this. Is that true?

#### romsek

MHF Helper
@romsek & JeffM: Thank you very much for the response. I have to apologize upfront because I neglected to state the textbook instructions: "Write a system of linear inequalities in two variables to represent each situation and graph. Determine one possible solution from the graph." The main concern I have here is how you choose the independent and dependent variable. It seems to me that it is arbitrary in a problem like this. Is that true?
The independent variable is the one you've got control over.

There are two here, oats and wheat.

The system would be

$oats + wheat \leq 540$

$2 oats \leq wheat$

The graph plots the total of oats and wheat for

oats:0 to it's max possible value of 180.

wheat: it's minimum value of 2(oats) to it's maximum value of 540-oats

#### Deveno

MHF Hall of Honor
You have here a system "bounded" by a linear equation:

wheat = twice oats.

One can think of this as the line $y = 2x$.

But this function is *invertible* (we can switch the $x$ and $y$ without restricting the domain), so we have an equivalent system bounded by the linear equation:

oats = half of wheat (which one can think of as the line $y = \frac{1}{2}x$).

To conform to "reality", one limits oneself to the first quadrant (some problems may not require this, some may have other minima/maxima inherent in their setup).

A final bound is the constraint oats + wheat = 540. Clearly in *this* bound, oats and wheat are on equal footing.

So, no, in this situation it does not matter which (oats or wheat) you make "dependent".

One final note: subject to his conservation concerns, MacGregor will probably want to consider a *yield* function, based on estimated profit (selling price/acre - cost/acre) of both oats and wheat (in other words, given the constraints of investment, one often seeks to maximize return).

basicmathguy1993

#### romsek

MHF Helper
You have here a system "bounded" by a linear equation:

wheat = twice oats.

One can think of this as the line $y = 2x$.

But this function is *invertible* (we can switch the $x$ and $y$ without restricting the domain), so we have an equivalent system bounded by the linear equation:

oats = half of wheat (which one can think of as the line $y = \frac{1}{2}x$).

To conform to "reality", one limits oneself to the first quadrant (some problems may not require this, some may have other minima/maxima inherent in their setup).

A final bound is the constraint oats + wheat = 540. Clearly in *this* bound, oats and wheat are on equal footing.

So, no, in this situation it does not matter which (oats or wheat) you make "dependent".

One final note: subject to his conservation concerns, MacGregor will probably want to consider a *yield* function, based on estimated profit (selling price/acre - cost/acre) of both oats and wheat (in other words, given the constraints of investment, one often seeks to maximize return).
I don't entirely agree. The problem suggests that not all 540 acres are necessarily used.

This results in both oats and wheat being independent with the single constraint that wheat > 2oats.

#### basicmathguy1993

Okay so there is a disagreement here. Deveno says that either wheats and oats can be dependent. The reason I'm confused is because the standard form of a linear equation in two variable, as you folks know, is ax + by = c. I just always assumed that y is always the dependent variable, because you can rearrange to slope-intercept form y = -(a/b)x + c/b. I did not know you could a linear equation in two variables with two independent variables.

#### romsek

MHF Helper
Okay so there is a disagreement here. Deveno says that either wheats and oats can be dependent. The reason I'm confused is because the standard form of a linear equation in two variable, as you folks know, is ax + by = c. I just always assumed that y is always the dependent variable, because you can rearrange to slope-intercept form y = -(a/b)x + c/b. I did not know you could a linear equation in two variables with two independent variables.
Do you have control over how much wheat you plant? Yes. It's an independent variable.

Do you have control over how much oats you plant? Yes. It too is an independent variable.

IF you are told that you must plant the entire 540 acres.

THEN oats+wheat=540, and the two are no longer independent. One can be determined from the other.

IF you are not told you must plant all 540 acres

THEN oats + wheat < 540, the two are now independent with a constraint on their relationship.

basicmathguy1993