Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
Concept Byte: Linear Programming
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Exercise 4 Page 163

Notice that all inequalities have to apply.

x=5 and y=9.5

Practice makes perfect

We are asked to find the values of x or y that maximize the given function. Notice that the fourth constraint, x≥ 0, limits the solution set to positive values of x. Let's start by isolating y in each inequality.

x+2y≤ 24 & (I) 3x+2y≤ 34 & (II) 3x+y≤ 29 & (III) x≥ 0 & (IV)
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(I), (II), (III): Write in slope-intercept form
2y≤ - x+ 24 3x+2y≤ 34 3x+y≤ 29 x≥ 0

(II), (III): LHS-3x=RHS-3x

2y≤ - x+ 24 2y≤ - 3x +34 y≤ -3x+29 x≥ 0

(I), (II): .LHS /2.=.RHS /2.

y≤ - x2+ 12 y≤ - 3x2 +17 y≤ -3x+29 x≥ 0

(I), (II): a* b/c=a/c* b

y≤ - 12x + 12 y≤ - 32x +17 y≤ -3x+29 x≥ 0

To solve the system of inequalities we need to enter them in the calculator. Push Y= and write the functions in the first two rows.

Next, we have to set the calculator to account for the inequality. Place the cursor before Y and scroll by pushing ENTER. We will select the icon that displays a black triangle pointing down and to the left. This tells the calculator that we are dealing with a ≤ inequality.

With the inequalities entered in the calculator, we can graph them by pushing GRAPH. Before we do that let's limit the window to x>0, as we have already established that we are interested in this area only. Make sure to adjust the viewing window so the y-intercepts are visible.

Notice that there are two points of intersection where all inequalities hold true. These have been pointed out above with a blue and red arrow. To determine which intersection maximizes the function we have to find both and compare. To do this, push CALC (2nd + TRACE) and choose the fifth option in the list, intersect.

Now we must provide the calculator with some guesses as to where the intersections occur. Below we have isolated the graphs that are of interest when calculating the different points of intersection.

To determine which point of intersection maximizes the objective function, we have to substitute both sets of coordinates into it and evaluate.

Coordinates 2x+3y P
x= 5, y= 9.5 2( 5)+3( 9.5) 38.5
x= 8, y= 5 2( 8)+3( 5) 31

Since 38.5>31, the coordinates that maximize the objective functions is x=5 and y=9.5.