ok here we are going to maximize the

function C equals x plus y given the constraints that why has to be greater

than or equal to 0 X has to be greater than or equal to 0 4x plus 2y must be

less than or equal to 8 and 2x minus y has to be less than or

equal to 0. Notice the first two constraints y greater than or equal to 0

and X greater than or equal to 0 that simply means that we’re going to be our

are feasible region is going to be in the first quadrant quadrant one so we’ll

keep that in mind and then i’m going to start graphing the inequalities 4x + 2 y

less than or equal to 8 and 2 X minus y less than or equal to 0 so I’m going to

start off by making both of these into equations so those these are just going

to be lines and the graph these i’m going to find the x and y-intercepts of

each so to find the x-intercept we can simply let y equal zero for the first

equation will be left with 4x equals eight if we divide both sides by 4 x

equals 2 for the second one well we’ll have 2x equals 0 if we divide both sides

by 2 will just get x equals 0 so this one’s a not going to be as useful

because well we’ll see in just a second to find the y-intercept we let X equal 0

and if we let X equal 0 we would have 2y equals 8 if we divide both sides by 2 will get y

equals four well ok back to what i was saying a

second ago if we let X equal 0 will have negative y equals 0 which tells us y

equals 0 so here we know we’re getting a point 2 comma 0 and 0 comma 4

on our line but for our second line we’re only getting the point 0 0 and

again we’re getting 00 so what we can do though 2x minus y equals 0 we could always

write that as 2 x equals y then we can graph it just using slope

intercept form ok so let’s start graphing here again we

said we’re in the first quadrant so I’m going to make my my graph so that we

just really only see the first quadrant ok so the first equation for x plus 2y

equals 8 there’s a point 2 comma 0 and also at 0 comma 4 against it was less

than or equal to the original constraint we can use solid to solid line and then

likewise we have y equals 2x so that’s going to have a y-intercept of 0 and

then we go up to over 1 up to over 1 etcetera ok so now we simply have to determine

the feasible region so what I’m going to do is use a test . again keeping in mind

that this has to be in the first quadrant so 4x plus 2y has to be less

than or equal to 8 and our other our other inequality was 2x minus y less

than or equal to 0 so for my line for x plus 2y equals 8 to test the inequality

i’m going to use a point not on that line and i met again just going to use

the origin 0 0 so again the other line does go through 00 but you can almost

forget about that line for a second so as long as I take a point off of the

line 4x plus 2y equals 8 I can use that as a test point so for my

first inequality i’m going to test the point 0 0 and certainly if you plug in 0

for X and 0 for y we would get that 0 is less than or equal 8 so I know for my first inequality i

would have to shade below for the second 12 X minus y less than or equal to 0 well now we can’t use the test point 00 so

we have to use some other point I don’t know how about the point 1 comma 0 so

I’m going to test the point 1 comma 0 so if we plug 1 in for X and 0 in for why i

have to ask myself does that satisfy our inequality well we’ll have to say is 2 less than

or equal to 0 certainly not so for my second inequality the test point was below

it which means we need to shade above it and again since we’re trapped in the first quadrant we know that our feasible region is going to be this

little triangular region it’s a touching the the y-axis and it’s easy to see that

this is going to cross at the point 1 comma 2 so that’s the benefit of having

a nice good graph ok so we can even label all the corner

points we’ve got 1 comma 2 we’ve got zero comma 0 the origin and then we also

have the point 0 comma 4 so now we simply need to take these three points

put them into the function that we were trying to maximize and determine which

of those gives us our largest value so we’ve got C equals we want to maximize c

equals x plus y and again we’ve got the points (0,0) (0,4) and the other one was 1

comma 2 well the arithmetic here is pretty easy i think so clearly if you

plug in 0 0 will get that c equals 0 0 plus 0 we put in 0 comma 4 will get that

c equals 0 + 4 or we will get the value of four and if we plug in 1 comma 2 into our

function will get 1 plus 2 way that only gives us three so it says the maximum

value possible will be the value of positive 4 and again that’s going to

occur at the point 0 comma 4

your videos have a tendency to be posted at the exactly perfect time, i have a midterm on this in 2 days. THANK YOU

I had one earlier today.. But still, thanks Patrick

you are great, thank you for this free resource.

i really like this channel. it helps me remember concepts that i have long forgotten.

Will you do a video explaining convolution please? I would greatly appreciate it!! Thanks!

I've tutored at a college and this is one of the subjects the students were learning. I loved helping them out with this and it was my first experience with Excel's solver. Btw I've tutored many, many students and I often recommend your videos. Keep up the great work Patrick!

I recommend him too when I tutor xD

patrick … the future is yours.

there is a faster way than doing the graphs

what rule rules out any other point along the triangle formed by the shaded area being the maximum? and i think its better to just use use the slope intercept form when deciding what to shade, that will leave you a y> (shade above) or y< (shade below), some thing but much easier to see i think

I farted during this video. ðŸ™‚

i love your handwriting. and nice tutorial. loved it!

I failed this class last semester.

Of course, my teacher didn't make it seem so easy, it was all about subsets and hyper planes and counting and inverse matrices that represented unimaginable geometric figures.

This stuff is so easy though. I can't but wonder why some people try and be dicks when explaining this.

Oh no, it's just that this guy was a (real) dick.

I had never complained about a teacher before, but this guy had a pretty questionable record; he told us that about 70% of his students dropped out from his class and that from the remaining 30%, only 40% of them passed.

I mean, I understand some teachers want to express that whatever they teach isn't easy, but I'm positive I can call this guy a dick without being unjustified.

thanks for this excellent technical aid, I surely send some moneys to my friend Patrick. Thanks Patrick, you are the best of the west and of the east too.

Will you ever do discrete math videos in the future? ðŸ™‚

Thankyou so much! That was really well explained!!

why must you go so fast?

seriously you make me want to bunch a baby in the throat

thank you!!!

Bro I just want to say I have watched a few of your videos and honestly they are really good, I now am going to take the time out to LIKE them thank you!! In a few Days I would like to post a video response about a difficult LP sum. Thank you

really love you videos wish you had some LP multiperiod word problems

Your video was a Godsent. Thank you!!!

Thanks so much life saver ðŸ™‚

I love you Patrick!!!! you always safe my life! :')

Your teaching is very good

It's all clear, but where do you get "test points"??????

gud teaching method…(y)

Excuse me how do you know whether to use the test points (0,0) or (1,0)? in other words how do I know which one to use and why didn't use (0,0) again for the second line?

how did you know how to graph y=2x?

Thank you! Appreciate the help.

So basically the function C we want to maximize is a plane in R^3. We could calculate the gradient of C to find in which direction its constant maximum slope occur and then draw equipotential lines to find at which corner the maximum value of C occurs

PATTY!!!! omfg I remember this channel why did it unsubscribe wtf YouTube

Upload more examples.

Hi Patrick, why couldn't test 0,0 be used again for the other line