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 1 Page 163

The and that minimize the function is the point of intersection of the boundary lines of the two first equations.

and

Practice makes perfect
We are asked to find the values of or that minimize the given function. Notice that the third set of constraints, and limit the solution set to the first quadrant. This means we should find the solution set of the first and second inequality in the first quadrant. Let's start by isolating in each inequality.
Write in slope-intercept form
To solve the system of inequalities we need to enter them in the calculator. Push 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 and scroll by pushing We will select the icon that displays a black triangle pointing up and to the right. This tells the calculator that we are dealing with a inequality.

With the inequalities entered in the calculator, we can graph them by pushing Before we do that let's limit the window to the first quadrant, as we have already established that we are interested in this area only.

The objective function is minimized where the boundary lines of the two first inequalities intersect. We can use the calculator to find this point of intersection. To do this, push CALC () and choose the fifth option in the list, intersect.

Now we must provide the calculator with some guesses as to where the intersection occurs.

The solution is and These values minimize the objective function.