quadrant I of a graph shown, 3 points are plotted at (0,8), (5,4), (9,0). a line goes through the points. the portion below the line is shaded.
Find the values of x and y that maximize the objective function P=3x+2y for the graph. What is the maximum value?
To find the values of x and y that maximize the objective function P=3x+2y, we need to look at the points within the shaded region below the line passing through the points (0,8), (5,4), and (9,0).
First, let's find the equation of the line passing through these points.
First, let's find the slope of the line using the points (0,8) and (9,0):
Slope (m) = (0 - 8) / (9 - 0) = -8/9
Then, we can find the y-intercept of the line using the point (0,8) and the slope:
y = mx + b
8 = -8/9 * 0 + b
b = 8
So, the equation of the line passing through the points is:
y = -8/9 x + 8
Now, we need to find the intersection points of the line y = -8/9 x + 8 and the x and y-axis.
Setting y = 0 in the equation of the line:
0 = -8/9 x + 8
8/9 x = 8
x = 9
So, the line intersects the x-axis at (9,0).
Setting x = 0 in the equation of the line:
y = -8/9 * 0 + 8
y = 8
So, the line intersects the y-axis at (0,8).
Now, we need to find the coordinates where the line intersects the boundary of the shaded region. This can be found by setting y = 0 in the equation of the line:
0 = -8/9 x + 8
8/9 x = 8
x = 9
So, the line intersects the boundary of the shaded region at (9,0).
Therefore, the values of x and y that maximize the objective function P = 3x + 2y are x = 9 and y = 0.
Substitute x = 9 and y = 0 into the objective function:
P = 3(9) + 2(0) = 27
Therefore, the maximum value of the objective function P is 27.