Exercise 26 Page 208

Introducing Variables

Let's start with introducing variables for the number of simple and grand arrangements the florist should make.

Writing Inequalities

We will use these variables to write inequalities representing the constraints given in the question.

Graphing the Inequalities

Let's graph the inequalities corresponding to the constraints.

Graphing 18g+10s≤ 2400

The graph of the linear inequality 18g+10s≤ 2400 is a half plane bounded by the line we get by replacing the inequality symbol with an equals sign. 18g+10s = 2400 We can graph this line by finding and connecting the intercepts.

• When g=0, then 10s=2400, so s= 240010=240.
• When s=0, then 18g=2400, so g= 240018=133.3.

Let's plot the intercepts and connect them with a line.

To see which half of the plane is to be shaded, we test a point. The florist needs 0+0=0 minutes to make 0 grand and 0 small arrangements. This is smaller than 2400 minutes, so the point (0,0) is in the shaded region.

Graphing 2g≤ s

The graph of the linear inequality 2g≤ s is a half plane bounded by the line we get by replacing the inequality symbol with equality. 2g=s We can graph this line by finding and connecting two of its points.

Let's plot these points and draw the line connecting them.

To see which half pane to shade, we test a point. The florist makes more simple arrangements than grand arrangements, so the point (100,100) is not in the shaded region.

Graphing g≥ 0 and s≥ 0

The inequalities g≥ 0 and s≥ 0 together mean that we only consider points in the first quadrant.

Graphing the Feasible Region

Let's now copy all graphs together on the same coordinate plane.

The feasible region is the overlapping part of the four graphs. We will need the vertices of this triangle, so let's highlight these on the graph.

Finding the Vertices

The maximum profit corresponds to one of the vertices, so let's find the coordinates. While graphing the boundary lines, we already found the coordinates of the vertices on the horizontal axis. (0,0) and (240,0) To find the third vertex, we need to find the intersection of the corresponding lines. 18g+10s=2400 2g=s Let's substitute s=2g in the first equation.

Notice that we rounded to the nearest integer because the florist can only make whole numbers of grand arrangements. Substituting this in the second equation gives the value of s. s=2g ⟹ s≈ 2(63)=126 Let's add this third point to our vertex list. (0,0) and (240,0) and (126,63)

Finding the Profit

We are given that the profit on a simple arrangement is $10 and the profit on a grand arrangement is $25. We can use these values to get an expression for the profit on s simple and g grand arrangements. P=10s+25g

Maximizing the Profit

We know that the profit is maximized at one of the vertices of the feasible region we found above. Let's find the profit in vertex (126,63).

We can find the profit at the other vertices similarly.

We can see that the maximum weekly profit the florist can achieve is $2835 when she makes and sells 126 simple and 63 grand arrangements.