20151128, 05:01  #1 
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
449 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 slopeintercept 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! 
20151128, 07:59  #2 
"Curtis"
Feb 2005
Riverside, CA
2·3^{2}·7·41 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? 
20151128, 08:42  #3  
May 2004
New York City
10213_{8} Posts 
Quote:


20151128, 11:34  #4  
Dec 2012
The Netherlands
6E1_{16} Posts 
Quote:
Quote:
The points to one side of the green line in your graph are the points with and the points on the other side are the points with . And if x=0 and y=0 then so the side containing the point (0,0) is the side for which . Thus the points in the graph for which are precisely the points on and to the left of the green line. Similarly, the points in the graph for which 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 with 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 . 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 so if we increase x by 1 then we have to decrease y by , which means that the value of 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 would therefore decrease. Thus the point with as big as possible is the point at which the two lines cross, and the value of 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 xy so we get the inequalities and , which simplify to When you draw the graph, you see immediately that the highest allowed value of x is the point at which the two lines and cross, and that value of x is the answer to the problem. I hope this helps! 

20151129, 01:37  #5 
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
449 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 20151129 at 01:40 Reason: Formatting 
20151129, 10:03  #6  
Dec 2012
The Netherlands
3×587 Posts 
Quote:
Just one small thing: you found y without using the fact that you already knew what x was. Once you know that , you can use that value in the equation to get immediately. The way you did it is correct, but this way involves fewer steps, allowing you to make the problem easier for yourself! 

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