# Math Help - URGENT! Drawing graph from linear programming

1. ## URGENT! Drawing graph from linear programming

Hi... i hope someone can help me because I'm sure i'm working this out correctly but these two lines should cross!!!! can anyone tell me if I'm going wrong anywhere...

Plotting 0.7c + 0.8t = 56
setting t=0
0.7c= 56
c= 56/0.7
c= 80
setting c=0
0.8t= 56
t= 56/0.8
t = 70

so with C on the y axis i have (0,80) and (70,0)

Plotting 0.4c + 0.5t =60
setting t=0
0.4c=60
c= 60/0.4
c= 150
setting c=0
0.5t=60
t= 60/0.5
t= 120
so with C on the y axis i have (0,150) and (120,0)

as you can see these two lines would not cross on a graph!!! these are definately the right numbers... unless i'm doing something wrong that i can't figure out i can only asume i've been given incorrect numbers!!

If anyone can shed some light Id be really greatful.. this is driving me nuts, its for some uni c/w due in VERY soon and there are about 25 marks of follw up question regarding feasle regions that i need the correct graph for!

2. Originally Posted by economics_student
Hi... i hope someone can help me because I'm sure i'm working this out correctly but these two lines should cross!!!! can anyone tell me if I'm going wrong anywhere...

Plotting 0.7c + 0.8t = 56
setting t=0
0.7c= 56
c= 56/0.7
c= 80
setting c=0
0.8t= 56
t= 56/0.8
t = 70

so with C on the y axis i have (0,80) and (70,0)

Plotting 0.4c + 0.5t =60
setting t=0
0.4c=60
c= 60/0.4
c= 150
setting c=0
0.5t=60
t= 60/0.5
t= 120
so with C on the y axis i have (0,150) and (120,0)

as you can see these two lines would not cross on a graph!!! these are definately the right numbers... unless i'm doing something wrong that i can't figure out i can only asume i've been given incorrect numbers!!

If anyone can shed some light Id be really greatful.. this is driving me nuts, its for some uni c/w due in VERY soon and there are about 25 marks of follw up question regarding feasle regions that i need the correct graph for!

What you're doing is correct! Why do you think these lines should cross? They often don't cross in a linear programming problem. It means one constraint, here the second one, is not effective. The feasible region will be formed by just one of the constraints and the non-negativity constraints (which I assume are there).

$
\setlength{\unitlength}{2.5cm}
\begin{picture}(1,1)(0,0)

\put(-.2,.75){80}
\put(.65,-.2){70}
\qbezier(0,0)(0,0)(0,.8)
\qbezier(0,0)(0,0)(.7,0)
\qbezier(0,.8)(0,.8)(.7,0)

\end{picture}
$

3. wow! ok... so the feasible region is within that line...

how would i work out the maximal profit then as i've only ever done it from where two lines cross? i.e. at corner points within the feasible region...

thank you so much for this by the way.. i've done thousands of simliar questions just to check that i'm not going crazy but they have all had lines which crossed!

4. Originally Posted by JakeD
What you're doing is correct! Why do you think these lines should cross? They often don't cross in a linear programming problem. It means one constraint, here the second one, is not effective. The feasible region will be formed by just one of the constraints and the non-negativity constraints (which I assume are there).

$
\setlength{\unitlength}{2.5cm}
\begin{picture}(1,1)(0,0)

\put(-.2,.75){80}
\put(.65,-.2){70}
\qbezier(0,0)(0,0)(0,.8)
\qbezier(0,0)(0,0)(.7,0)
\qbezier(0,.8)(0,.8)(.7,0)

\end{picture}
$
Originally Posted by economics_student
wow! ok... so the feasible region is within that line...

how would i work out the maximal profit then as i've only ever done it from where two lines cross? i.e. at corner points within the feasible region...
The maximal profit point will still be at a corner of the feasible region. The second constraint just drops out. The graph might look like this:

$
\setlength{\unitlength}{2.5cm}
\begin{picture}(1,1)(0,0)

\put(-.2,.75){80}
\put(.65,-.2){70}
\qbezier(0,0)(0,0)(0,.8)
\qbezier(0,0)(0,0)(.7,0)
\qbezier(0,.8)(0,.8)(.7,0)
\qbezier(0,.8)(0,.8)(.8,.25)

\end{picture}
$

It's just that previously in order to find the optimal solution i have simultaneously solved the equations of the two line which crossed at the corner point, i'm not sure how I'd do this in this example.

Thank you again.

6. Originally Posted by JakeD
The maximal profit point will still be at a corner of the feasible region. The second constraint just drops out. The graph might look like this:

$
\setlength{\unitlength}{2.5cm}
\begin{picture}(1,1)(0,0)

\put(-.2,.75){80}
\put(.65,-.2){70}
\qbezier(0,0)(0,0)(0,.8)
\qbezier(0,0)(0,0)(.7,0)
\qbezier(0,.8)(0,.8)(.7,0)
\qbezier(0,.8)(0,.8)(.8,.25)

\end{picture}
$
Originally Posted by economics_student

It's just that previously in order to find the optimal solution i have simultaneously solved the equations of the two line which crossed at the corner point, i'm not sure how I'd do this in this example.

Thank you again.
A linear program is solved by

• computing all the corner points of the feasible region,
• evaluating the objective function at those corner points,
• selecting the corner point with the highest value for the objective function.

In this example, the feasible region has 3 corner points: (0,80), (70,0) and (0,0). Evaluate the objective function at those points and select the point with the highest value.

In finding the corner points here, you will be using one or both non-negativity constraints $c \ge 0,\ t \ge 0$ for your lines. For example, for the corner point $(t,c) = (0,80),$ the two lines are $0.7c + 0.8t = 56$ and $t = 0$ from the constraints $0.7c + 0.8t \le 56$ and $t \ge 0.$

7. ok... so the max profit in this question is
6c+10t
so basically its a choice of:
6(80) + 10(0) = 480
or
6(0) + 10(70) = 700

so the optimal production combination is c=0 and t=70
with a profit contribution of 700 (?)

8. Originally Posted by economics_student
ok... so the max profit in this question is
6c+10t
so basically its a choice of:
6(80) + 10(0) = 480
or
6(0) + 10(70) = 700

so the optimal production combination is c=0 and t=70
with a profit contribution of 700 (?)
Yes. And the graph with the iso-profit line $6c + 10t = 700$ drawn in looks like this:

$
\setlength{\unitlength}{2.5cm}
\begin{picture}(1,1)(0,0)

\put(-.2,.75){80}
\put(.65,-.2){70}
\qbezier(0,0)(0,0)(0,.8)
\qbezier(0,0)(0,0)(.7,0)
\qbezier(0,.8)(0,.8)(.7,0)
\qbezier(.7,0)(.7,0)(.1,1)

\end{picture}
$

9. You are truely amazing!

Thank you so much... I really have been going over this for a long time now! (about 6 days to be exact!!)

All of our previous excercises have been with crossing lines- i just wasn't thinking outside the box! it was very sneaky of them!

Thank you again you have helped so much!!!!

10. Originally Posted by economics_student
Hi... i hope someone can help me because I'm sure i'm working this out correctly but these two lines should cross!!!! can anyone tell me if I'm going wrong anywhere...

Plotting 0.7c + 0.8t = 56
so with C on the y axis i have (0,80) and (70,0)

Plotting 0.4c + 0.5t =60
so with C on the y axis i have (0,150) and (120,0)

as you can see these two lines would not cross on a graph!!! these are definately the right numbers... unless i'm doing something wrong that i can't figure out i can only asume i've been given incorrect numbers!!

The given information is not enough.

Sure the two lines will cross each other since they are not parallel. The intersection point is at (653.33,-666.67). Not in the first quadrant, but in the 4th quadrant---away from the feasible region.

The two constraints are in equation forms? Not in inequalities?
By plotting the two lines on the same (t,c) rectangular axes, a quadrilateral with corner points (0,80), (0,150), (120,0) and (70,0) is formed. I assume then that the two constraints are supposed to be
0.7C +0.8t >= 56 ------(1) ...."greater than, or equal to"
0.4c +0.5t <= 60 ------(2) ..."less than, or equal to"
And so the feasible region is the said quadrilateral.

Then test each of the four corner points against the objective function to find which one gives the optimal objective.

11. Hello, economics_student!

I suspect that there is a typo in the problem.
. . The lines do intersect, but in Quadrant 2.

$\begin{array}{ccc}0.7x + 0.8y & = & 56 \\
0.4x + 0.5y & =& 60\end{array}$

Multiply both equatons by 10: . $\begin{array}{cccc}7x + 8y & = & 560 & [1] \\ 4x + 5y & = & 600 & [2]\end{array}$

$\begin{array}{cccc}\text{Mutliply [1] by 5:} & 35x + 40y & = & 2800 \\ \text{Multiply [2] by -8:} & \text{-}32x - 40y & = & \text{-}4800\end{array}$

. . Add: . $3x \:=\:\text{-}2000\quad\Rightarrow\quad \boxed{x \:=\:-\frac{2000}{3}}$

Substitute into [2]: . $4\left(-\frac{2000}{3}\right) + 5y \:=\:600\quad\Rightarrow\quad\boxed{y \:=\:\frac{1960}{3}}$

12. Originally Posted by economics_student
Hi... i hope someone can help me because I'm sure i'm working this out correctly but these two lines should cross!!!! can anyone tell me if I'm going wrong anywhere...

Plotting 0.7c + 0.8t = 56
setting t=0
0.7c= 56
c= 56/0.7
c= 80
setting c=0
0.8t= 56
t= 56/0.8
t = 70

so with C on the y axis i have (0,80) and (70,0)

Plotting 0.4c + 0.5t =60
setting t=0
0.4c=60
c= 60/0.4
c= 150
setting c=0
0.5t=60
t= 60/0.5
t= 120
so with C on the y axis i have (0,150) and (120,0)

as you can see these two lines would not cross on a graph!!! these are definately the right numbers... unless i'm doing something wrong that i can't figure out i can only asume i've been given incorrect numbers!!

If anyone can shed some light Id be really greatful.. this is driving me nuts, its for some uni c/w due in VERY soon and there are about 25 marks of follw up question regarding feasle regions that i need the correct graph for!

Originally Posted by JakeD
What you're doing is correct! Why do you think these lines should cross? They often don't cross in a linear programming problem. It means one constraint, here the second one, is not effective. The feasible region will be formed by just one of the constraints and the non-negativity constraints (which I assume are there).

$
\setlength{\unitlength}{2.5cm}
\begin{picture}(1,1)(0,0)

\put(-.2,.75){80}
\put(.65,-.2){70}
\qbezier(0,0)(0,0)(0,.8)
\qbezier(0,0)(0,0)(.7,0)
\qbezier(0,.8)(0,.8)(.7,0)

\end{picture}
$
Originally Posted by ticbol
The given information is not enough.

Sure the two lines will cross each other since they are not parallel. The intersection point is at (653.33,-666.67). Not in the first quadrant, but in the 4th quadrant---away from the feasible region.

The two constraints are in equation forms? Not in inequalities?
By plotting the two lines on the same (t,c) rectangular axes, a quadrilateral with corner points (0,80), (0,150), (120,0) and (70,0) is formed. I assume then that the two constraints are supposed to be
0.7C +0.8t >= 56 ------(1) ...."greater than, or equal to"
0.4c +0.5t <= 60 ------(2) ..."less than, or equal to"
And so the feasible region is the said quadrilateral.

Then test each of the four corner points against the objective function to find which one gives the optimal objective.
Originally Posted by Soroban
Hello, economics_student!

I suspect that there is a typo in the problem.
. . The lines do intersect, but in Quadrant 2.

Multiply both equatons by 10: . $\begin{array}{cccc}7x + 8y & = & 560 & [1] \\ 4x + 5y & = & 600 & [2]\end{array}$

$\begin{array}{cccc}\text{Mutliply [1] by 5:} & 35x + 40y & = & 2800 \\ \text{Multiply [2] by -8:} & \text{-}32x - 40y & = & \text{-}4800\end{array}$

. . Add: . $3x \:=\:\text{-}2000\quad\Rightarrow\quad \boxed{x \:=\:-\frac{2000}{3}}$

Substitute into [2]: . $4\left(-\frac{2000}{3}\right) + 5y \:=\:600\quad\Rightarrow\quad\boxed{y \:=\:\frac{1960}{3}}$

ticbol and Soroban have raised the question of what the actual problem given was. I agree with that question and I should have asked it. Instead, I assumed the following.

\begin{aligned}
0.7c +0.8t &\le 56 \\
0.4c +0.5t &\le 60 \\
c \ge 0,&\ t \ge 0
\end{aligned}

This is not an unusual linear programming problem. The graph of the constraints looks like the one below and the feasible region is the inner triangle. The "lines don't intersect" means in the first quadrant, which is the only relevant quadrant because of the non-negativity constraints. This is a case of the second constraint being completely ineffective. When you learn about linear programming, you should learn that this can happen. But, again, I agree it is not clear this was the actual problem given and I should have asked about it.

$
\setlength{\unitlength}{2.5cm}
\begin{picture}(1,1)(0,0)

\put(-.2,.75){80}
\put(.65,-.2){70}
\put(-.3,1.45){150}
\put(1.15,-.2){120}
\qbezier(0,0)(0,0)(0,1.5)
\qbezier(0,0)(0,0)(1.2,0)
\qbezier(0,.8)(0,.8)(.7,0)
\qbezier(0,1.5)(0,1.5)(1.2,0)

\end{picture}
$