### Video Transcript

Find the maximum value of the
objective function π equals two π₯ plus six π¦ given the constraints π₯ is greater
than or equal to zero, π¦ is greater than or equal to zero, π₯ plus π¦ is less or
equal to six, three π₯ plus π¦ is less than or equal to nine, and π₯ plus two π¦ is
less than or equal to eight.

One of these shaded regions
actually is where all of these constraints overlap. Letβs go ahead and just figure out
ourselves which of these shaded regions represents that. Our first constraint is π₯ is
greater than or equal to zero, which we can see here. Itβs a solid line and everything
greater than zero, so above that line. π¦ is greater than or equal to
zero; we can see here everything to the right of that line cause theyβre positive
values and itβs a solid line.

Now, our third constraint, we can
rewrite that in the form of π¦ equals ππ₯ plus π, but using an inequality
sign. By subtracting π₯ from both sides,
we get π¦ is less than or equal to negative π₯ plus six, which means we should cross
the π¦-axis at six and have a slope of negative one, which is this line. And everything less than that would
be below that.

Our fourth constraint, we can
subtract three π₯ from both sides and get that π¦ is less than or equal to negative
three π₯ plus nine. So we cross the π¦-axis at nine and
have a slope of negative three. So here we can see this inequality
on our graph. And then lastly, so our last
inequality can be represented by π¦ is less than or equal to negative π₯ over two
plus four because we will subtract both sides by π₯ and then divide everything by
two. So this inequality will cross the
π¦-axis at four and have a slope of negative one-half.

And here we can see it on our
graph. So where all of these constraints
overlap would be this region. So to find the maximum value of our
function, we need to check all of the vertexes: so zero, four, two, three, three,
zero, and then our last vertex of zero, zero. Thatβs definitely not going to give
us a maximum value because if weβll plug in zero for π₯ and zero for π¦, weβre just
gonna get zero. So we can skip that one.

So letβs take each of these points,
plug it into our objective function, and see which one gives us the highest
value. So plugging in zero for π₯ and four
for π¦, we have two times zero plus six times four, so zero plus 24, which is
24. Now, we plug in two, three. So two times two plus six times
three, so four plus 18, which gives us 22. And now lastly, we plug in three,
zero. So two times three plus six times
zero, so six plus zero is six. Therefore, 24 is the maximum
value.