Find the maximum value of the objective function 𝑝 equals two 𝑥 plus six 𝑦 given
the constraints 𝑥 is greater than or equal to zero, 𝑦 is greater than or equal to zero, 𝑥 plus 𝑦
is less or equal to six, three 𝑥 plus 𝑦 is less than or equal to nine, and 𝑥 plus two 𝑦 is less
than or equal to eight.
One of these shaded regions actually is where all of these constraints
overlap. Let’s go ahead and just figure out ourselves which of these shaded regions
represents that. Our first constraint is 𝑥 is greater than or equal to zero, which we can see
here. It’s a solid line and everything greater than zero, so above that line. 𝑦 is greater than
or equal to zero; we can see here everything to the right of that line cause they’re positive
values and it’s a solid line.
Now, our third constraint, we can rewrite that in the form of 𝑦
equals 𝑚𝑥 plus 𝑏, but using an inequality sign. By subtracting 𝑥 from both sides, we get 𝑦 is
less than or equal to negative 𝑥 plus six, which means we should cross the 𝑦-axis at six and
have a slope of negative one, which is this line. And everything less than that would be below
Our fourth constraint, we can subtract three 𝑥 from both sides and get that 𝑦 is less than
or equal to negative three 𝑥 plus nine. So we cross the 𝑦-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 𝑦 is less than or equal to negative 𝑥 over two plus four
because we will subtract both sides by 𝑥 and then divide everything by two. So this inequality
will cross the 𝑦-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 vertices: so zero, four, two, three, three, zero,
and then our last vertice [vertex] of zero, zero. That’s definitely not going to give us a maximum
value because if we’ll plug in zero for 𝑥 and zero for 𝑦, we’re just gonna get zero. So we can
skip that one.
So let’s 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 𝑥 and four for 𝑦, 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.