Finding line intersection values?

jvang · 2015-11-28, 05:01

I'm currently working on some algebra problems (nothing hard) and had some trouble with this problem.

I have 2 nonstrict inequalities on a graph, but they are using fractions and my graph is in integers. How would I find the maximum of these two inequalities, or where they cross? I feel really dumb because I used to know how to do this. I think it would involve using the lines' equations...?

This is the problem:
I have 50 units of asphalt and 80 units of paint. To build one mile of highway on land, I use 2 units of asphalt and 1 unit of paint. To build 1 mile of highway on bridges, I use 1 unit of asphalt and 3 units of paint. What is the maximum number of miles of highway I can build?

$x$ = land mile resources
$y$ = bridge mile resources

Maximum units of asphalt: $2x + y \leq 50$; in slope-intercept form: $y \leq 50 - 2x$

Maximum units of paint: $x + 3y \leq 80$; in slope intercept form: $y \leq \frac{26}{3} - \frac{1}{3}x$

I then graphed these inequalities and got stuck. I have attached a rough representation below.

Thanks for your help!

VBCurtis · 2015-11-28, 07:59

What does the intersection point on the graph represent?

If you change the inequalities to equations, what does the solution of the system achieve?

davar55 · 2015-11-28, 08:42

Quote:

Originally Posted by jvang

I'm currently working on some algebra problems (nothing hard) and had some trouble with this problem.
I have 2 nonstrict inequalities on a graph, but they are using fractions and my graph is in integers. How would I find the maximum of these two inequalities, or where they cross? I feel really dumb because I used to know how to do this. I think it would involve using the lines' equations...?
This is the problem:
I have 50 units of asphalt and 80 units of paint. To build one mile of highway on land, I use 2 units of asphalt and 1 unit of paint. To build 1 mile of highway on bridges, I use 1 unit of asphalt and 3 units of paint. What is the maximum number of miles of highway I can build?
$x$ = land mile resources
$y$ = bridge mile resources
Maximum units of asphalt: $2x + y \leq 50$; in slope-intercept form: $y \leq 50 - 2x$
Maximum units of paint: $x + 3y \leq 80$; in slope intercept form: $y \leq \frac{26}{3} - \frac{1}{3}x$
I then graphed these inequalities and got stuck. I have attached a rough representation below.
Thanks for your help!

Just an arithmetic error: the equation x+3y <= 80 leads to y = (26 + 2/3) - (1/3)x. Then regraph, or solve the simultaneous equations. If the result is non-integral, the actual best integral solution is nearby.

Nick · 2015-11-28, 11:34

Quote:

Originally Posted by jvang

I have 2 nonstrict inequalities on a graph, but they are using fractions and my graph is in integers.

Don't worry about the fractions - in a problem like this, you may have to use fractions even when all the numbers in the problem itself (and the answer) are integers.

Quote:

Originally Posted by jvang

How would I find the maximum of these two inequalities, or where they cross? ...I think it would involve using the lines' equations...?

Absolutely right. The green line in your graph gives all the points $(x,y)$ for which $y=50-2x$ and the purple line gives all the points $(x,y)$ for which $y=\frac{80}{3}-\frac{x}{3}$ so the point at which they cross is the point $(x,y)$ satisfying both these equations. At that point, we therefore have $50-2x=\frac{80}{3}-\frac{x}{3}$ (both expressions give the value of y at that point, so they must be equal). From this, you can work out the value of x where the lines cross, and then you can use either equation to work out the value of y where the lines cross, too.

The points to one side of the green line in your graph are the points $(x,y)$ with $2x+y<50$ and the points on the other side are the points $(x,y)$ with $2x+y>50$ . And if x=0 and y=0 then $2x+y<50$ so the side containing the point (0,0) is the side for which $2x+y<50$ . Thus the points $(x,y)$ in the graph for which $2x+y \leq 50$ are precisely the points on and to the left of the green line. Similarly, the points $(x,y)$ in the graph for which $x+3y \leq 80$ are precisely the points on and underneath the purple line. It follows that the points satisfying both inequalities are precisely those points that are both to the left of or on the green line and underneath or on the purple line (this is the region in the bottom left of your graph). Out of these points, we want to find a point $(x,y)$ with $x+y$ as big as possible.

This is where it gets a bit tricky with the meanings you chose for x and y. So let's finish working it out this way, but then consider how you could make it easier for yourself. In the region whose points satisfy both inequalities, the point with the biggest value of y is $(x,y)=(0,\frac{80}{3})$ . If we try to increase x, we must decrease y in order to stay underneath the purple line. And the slope of the purple line is $-\frac{1}{3}$ so if we increase x by 1 then we have to decrease y by $\frac{1}{3}$ , which means that the value of $x+y$ has increased, so this is worth doing. Thus, we move along to the point where the 2 lines of the graph cross (the point calculated above). If we want to increase x any more, we are now limited by the green line. And the slope of the green line is -2 so if we increase x further then the value of y will go down faster than the value of x goes up, and the value of $x+y$ would therefore decrease. Thus the point $(x,y)$ with $x+y$ as big as possible is the point at which the two lines cross, and the value of $x+y$ at that point is the answer to the problem.

The part above is difficult because the value we want to make as big as possible is not x or y but x+y. If we start the whole problem again but writing x for the number of miles of highway both on land and on bridges and y for the number of miles of highway on land, then the number of miles of highway on bridges is x-y so we get the inequalities $2y+(x-y)\leq 50$ and $y+3(x-y)\leq 80$ , which simplify to
$\begin{eqnarray*}<br /> x+y & \leq & 50 \\<br /> 3x-2y & \leq & 80<br /> \end{eqnarray*}$
When you draw the graph, you see immediately that the highest allowed value of x is the point at which the two lines $x+y=50$ and $3x-2y=80$ cross, and that value of x is the answer to the problem.

I hope this helps!

jvang · 2015-11-29, 01:37

Thanks for the help! I finished up the problem, which is detailed below.

To find x:

$y \leq 50 - 2x\\

y \leq \frac{80}{3} - \frac{1}{3}x\\

50 - 2x = \frac{80}{3} - \frac{1}{3}x\\

*3\\

150 - 6x = 80 - x\\

+6x\\

150 = 80 + 5x\\

-80\\

70 = 5x\\

\div 5\\

x = 14$

Using the original equations to find y:

$2x + y \leq 50\\

-y\\

2x \leq 50 - y\\

\div 2\\

x \leq 25 - \frac{1}{2} y\\

x + 3y \leq 80\\

-3y\\

x \leq 80 - 3y\\

25 - \frac{1}{2} y = 80 - 3y\\

*2\\

50 - y = 160 - 6y\\

+6y\\

50 + 5y = 160\\

-50\\

5y = 110\\

\div 5\\

y = 22\\

x + y\\

14 + 22\\$

$\fbox{36}$ is the maximum.

Is it necessary to show all of the steps or is it already easy to follow? I wasn't sure, so I went ahead and included it. Also, how presentable is my $\small\LaTeX$?

Thanks again!

Nick · 2015-11-29, 10:03

Quote:

Originally Posted by jvang

$\fbox{36}$ is the maximum.

Is it necessary to show all of the steps or is it already easy to follow? I wasn't sure, so I went ahead and included it. Also, how presentable is my $\small\LaTeX$ ?

That's great - you have found the correct answer (and your $\small\LaTeX$ is fine).

Just one small thing: you found y without using the fact that you already knew what x was.
Once you know that $x=14$ , you can use that value in the equation $y=50-2x$ to get $y=22$ immediately.
The way you did it is correct, but this way involves fewer steps, allowing you to make the problem easier for yourself!

2015-11-28, 05:01	#1
jvang veganjoy "Joey" Nov 2015 Middle of Nowhere,AR 3×5×29 Posts	Finding line intersection values? I'm currently working on some algebra problems (nothing hard) and had some trouble with this problem. I have 2 nonstrict inequalities on a graph, but they are using fractions and my graph is in integers. How would I find the maximum of these two inequalities, or where they cross? I feel really dumb because I used to know how to do this. I think it would involve using the lines' equations...? This is the problem: I have 50 units of asphalt and 80 units of paint. To build one mile of highway on land, I use 2 units of asphalt and 1 unit of paint. To build 1 mile of highway on bridges, I use 1 unit of asphalt and 3 units of paint. What is the maximum number of miles of highway I can build? \(x\) = land mile resources \(y\) = bridge mile resources Maximum units of asphalt: \(2x + y \leq 50\); in slope-intercept form: \(y \leq 50 - 2x\) Maximum units of paint: \(x + 3y \leq 80\); in slope intercept form: \(y \leq \frac{26}{3} - \frac{1}{3}x\) I then graphed these inequalities and got stuck. I have attached a rough representation below. Thanks for your help! Attached Thumbnails

2015-11-29, 01:37	#5
jvang veganjoy "Joey" Nov 2015 Middle of Nowhere,AR 1B3₁₆ Posts	Thanks for the help! I finished up the problem, which is detailed below. To find x: \(y \leq 50 - 2x\\ y \leq \frac{80}{3} - \frac{1}{3}x\\ 50 - 2x = \frac{80}{3} - \frac{1}{3}x\\ 3\\ 150 - 6x = 80 - x\\ +6x\\ 150 = 80 + 5x\\ -80\\ 70 = 5x\\ \div 5\\ x = 14\) Using the original equations to find y: \(2x + y \leq 50\\ -y\\ 2x \leq 50 - y\\ \div 2\\ x \leq 25 - \frac{1}{2} y\\ x + 3y \leq 80\\ -3y\\ x \leq 80 - 3y\\ 25 - \frac{1}{2} y = 80 - 3y\\ 2\\ 50 - y = 160 - 6y\\ +6y\\ 50 + 5y = 160\\ -50\\ 5y = 110\\ \div 5\\ y = 22\\ x + y\\ 14 + 22\\\) \(\fbox{36}\) is the maximum. Is it necessary to show all of the steps or is it already easy to follow? I wasn't sure, so I went ahead and included it. Also, how presentable is my \(\small\LaTeX\)? Thanks again! Last fiddled with by jvang on 2015-11-29 at 01:40 Reason: Formatting

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2015-11-28, 07:59	#2
VBCurtis "Curtis" Feb 2005 Riverside, CA 5²·11·17 Posts	What does the intersection point on the graph represent? If you change the inequalities to equations, what does the solution of the system achieve?