Exercise 24 Page 158

To find the number of brochures and fliers that minimize the cost, we will follow three steps.

We should first write the constraints.
Then, we will graph the feasible region and find its vertices.
Finally, we will write the objective function and use the vertices to find the minimum cost.

Let's start!

Constraints

Let x be the number of brochures and y be the number of fliers. The manager needs at least 50 brochures and 150 fliers. With this, first two constraints can be written. x≥ 50 & (I) y≥ 150 & (II) For the third constraint, we will use the given data and make an organized table to write it.

Verbal Expression	Algebraic Expression
Number of pages for x number of brochures	3 x
Number of pages for y number of fliers	2 y
Total number of pages is less than or equal to 600 hours.	3 x+ 2 y≤ 600

As we can see, we have three constraints to write a system. x≥ 50 & (I) y≥ 150 & (II) 3x+2y ≤ 600 & (III)

Feasible Region

To graph the constraints, we should first determine their boundary lines. They can be determined by replacing the inequality symbol with the equals sign. Let's begin with Inequality I and Inequality II.

ccc & &Inequality &Boundary Line & I: &x ≥ 50 &x = 50 & II: &y ≥ 150 &y = 150 Boundary Line I is a vertical line that passes through the point (50,0). Boundary Line II is a horizontal line that passes through the point (0,150). Because of the non-strict inequalities, the boundary lines will be solid.

Inequality I states that the points with x-coordinates greater than or equal to 50 are included in the solution. Therefore, we will shade the region to right of Boundary Line I. With the same reasoning, we will shade the region above Boundary Line II.

Next, we will graph Inequality III. Let's first determine its boundary line. cc &Inequality III &Boundary Line III &3x+2y ≤ 600 &3x+2y = 600 The boundary line is in standard form. Therefore, it would be a better option to find its intercepts to graph it. We will substitute y= 0 for the x-intercept and x= 0 for the y-intercept.

	3x+2y = 600
Operation	x-intercept	y-intercept
Substitution	3x+2( 0) = 600	3( 0)+2y = 600
Calculation	x=200	y=300
Point	(200,0)	(0,300)

Now we can plot the intercepts and connect them with a line segment. Notice that the number of brochures and fliers cannot be negative, so the line will be bound by the axes. The boundary line will also be solid because of the non-strict inequality.

Next, we will decide which region we should shade by testing the point (0,0).

3x+2y ≤ 600

SubstituteII

x= 0, y= 0

3( 0)+2( 0) ? ≤ 600

ZeroPropMult

Zero Property of Multiplication

0≤ 600

Since the point satisfies the inequality, region that contains the point will be shaded. Let's do it!

The overlapping section of the graph above represents the feasible region. The points of intersection are the vertices of the feasible region.

Vertices (50,150), (50,225), (100,150)

Objective Function and Minimum Cost

Let C be the total cost. We will make an organized table to write a function for the total cost.

Verbal Expression	Algebraic Expression
Cost of x number of brochures ($)	0.08 x
Cost of y number of fliers ($)	0.04 y
Total cost is $C.	C= 0.08 x+ 0.04 y

To find the number of brochures and fliers that minimize the cost, we will substitute the vertices into the function.

Vertex	0.08x+0.04y	C
( 50, 150)	0.08( 50)+0.04( 150)	$10
( 50, 225)	0.08( 50)+0.04( 225)	$13
( 100, 150)	0.08( 100)+0.04( 150)	$14

As a result, to minimize the cost, 50 brochures and 150 fliers should be printed. With this, the minimum cost will be $10.

Hints

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Practice exercises

Asses your skill

Constraints

Feasible Region

Objective Function and Minimum Cost