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P=12 is maximized at (0,4)
Our first step in finding the maximum value of the given equation is to graph the system and determine the vertices of the overlapping region. Substituting these vertices into the equation, we will find the maximum value. x+y≤ 5 & (I) x+2y ≤ 8 & (II) x≥ 0,y≥ 0 & (III)
Since the point satisfies the inequality, we will shade the region that contains the point.
Notice that for the third inequality the boundary lines are x=0 and y=0. Therefore, they overlap with the x- and y-axis.
Our inequalities are x≥ 0 and y≥ 0. This means that every coordinate pair with x- and y-value greater than or equal to 0 needs to be included in the shaded region. This indicates that we should shade the first quadrant.
Let's draw the graphs of the inequalities on the same coordinate plane.
Now that we can see the overlapping region, let's highlight the vertices.
x= 0, y= 0
Zero Property of Multiplication
Add terms
Vertex | P=x+3y | Value |
---|---|---|
( 0, 0) | P= 0+3( 0) | P=0 |
( 5, 0) | P= 5+3( 0) | P=5 |
( 0, 4) | P= 0+3( 4) | P=12 |
( 2, 3) | P= 2+3( 3) | P=11 |
Looking at the table, we can see that the maximum value of the equation is P=12.