# Question Video: Finding the Maximum Value of the Objective Function given the Constraints and Their Graph Mathematics • 9th Grade

Find the maximum value of the objective function π = 2π₯ + 6π¦ given the constraints π₯ β₯ 0, π¦ β₯ 0, π₯ + π¦ β€ 6, 3π₯ + π¦ β€ 9, and π₯ + 2π¦ β€ 8.

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### 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. Letβs 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. Itβs 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 theyβre positive values and itβs a solid line.

Now, our third constraint, we can rewrite that in the form of π¦ equals ππ₯ 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. Thatβs definitely not going to give us a maximum value because if weβll plug in zero for π₯ and zero for π¦, weβre just gonna get zero. So we can skip that one.

So letβ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 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.