Exercise 13 Page 160

a We are given a table that represents the Spruce and Maple Tree Data.

	Spruce	Maple
Planting Cost	$$ 30$	$$ 40$
Area Required	$600 ft^{2}$	$900 ft^{2}$
Carbon Dioxide Absorption	$650$ lb/yr	$300$ lb/yr

Let

x

be the number of spruces and

y

be the number of maples. We can write the first two constraints based on the fact that the number of trees cannot be negative.

Constraint I : Constraint II : x \geq 0 y \geq 0

Since the city has

$ 2100

to spend on planting, the total cost of the trees will be less than or equal to the city's budget.

\underline{Constraint III} 30 x + 40 y \leq 2100 \Leftrightarrow 3 x + 4 y \leq 210

The available area for planting in the city is

45000 ft^{2} .

Therefore, the total area covered by the trees will be less than or equal to city's available area.

\underline{Constraint IV} 600 x + 900 y \leq 45000 \Leftrightarrow 2 x + 3 y \leq 150

b The quantity we are trying to maximize or minimize is modeled with an objective function. In this case, we are trying to maximize the carbon dioxide absorption. Therefore, the objective function can be written using the third row of the table.

Objective Function
Verbal Expression	Algebraic Expression
$CO_{2}$ absorbed by Spruce trees (lb/yr)	$650 x$
$CO_{2}$ absorbed by Maple trees (lb/yr)	$300 y$
Total $CO_{2}$ absorbed by the trees (lb/yr)	$650 x + 300 y$
Total $CO_{2}$ absorbed by the trees is $A (lb / yr)$	$A = 650 x + 300 y$

c Let's start with the first two constraints. We will first determine their boundary lines by replacing the inequality sign with the equals sign.

I : II : \underline{Constraint} x \geq 0 y \geq 0 \underline{Boundary Line} x = 0 y = 0

The boundary lines of the first two constraints are the axes. Since both

x

and

y

are greater than or equal to

0,

we will shade Quadrant I. Notice that the inequalities are non-strict, so the boundary lines will be solid.

Next, we will graph Constraint III and IV by finding their boundary lines.

III : IV : \underline{Constraint} 3 x + 4 y \leq 210 2 x + 3 y \leq 150 \underline{Boundary Line} 3 x + 4 y = 210 2 x + 3 y = 150

To graph their boundary lines, we will find the

x -

and

y -

intercepts of the lines. We will first substitute

0

for

x

into Boundary Line III and solve it for

y

to find its

y -

intercept.

3 x + 4 y = 210

Substitute

$x = 0$

3 (0) + 4 y = 210

ZeroPropMult

Zero Property of Multiplication

4 y = 210

DivEqn

$LHS / 4 = RHS / 4$

y = 52.5

The

y -

intercept of Boundary Line III is the point

(0, 52.5) .

Proceeding in the same way, we can find the intercepts of both lines.

Constraint	Boundary Line	$x = 0$	$y -$ intercept	$y = 0$	$x -$ intercept
III	$3 x + 4 y = 210$	$3 (0) + 4 y = 210$	$(0, 52.5)$	$3 x + 4 (0) = 210$	$(70, 0)$
IV	$2 x + 3 y = 150$	$2 (0) + 3 y = 150$	$(0, 50)$	$2 x + 3 (0) = 150$	$(75, 0)$

Now we will plot the intercepts and connect them with line segments. The lines will be bound by the axes because the number of trees cannot be negative. Also, notice that the constraints are non-strict. Therefore, the lines will be solid.

Next, we will test the point

(0, 0)

to determine which region we should shade. Let's start with Constraint III.

3 x + 4 y \leq 210

SubstituteII

$x = 0$ , $y = 0$

3 (0) + 4 (0) \leq ? 210

ZeroPropMult

Zero Property of Multiplication

0 \leq 210

For Constraint III, we will shade the region that contains the point.

Now we will test the point for Constraint IV.

2 x + 3 y \leq 150

SubstituteII

$x = 0$ , $y = 0$

2 (0) + 3 (0) \leq ? 150

ZeroPropMult

Zero Property of Multiplication

0 \leq 150

Therefore, we will again shade the region that contains the test point.

The overlapping section of the graph above is the feasible region. Let's remove the unnecessary parts and indicate the vertices of the feasible region.

The coordinates of three of the four vertices are determined. The fourth vertex is the point of intersection of Constraint III and IV. To determine it we will first write a system of equations.

{3 x + 4 y = 210 2 x + 3 y = 150

To solve the system, we will equate the coefficients of

x

and use the Elimination Method.

{3 x + 4 y = 210 2 x + 3 y = 150

MultEqn

$(I):$ $LHS \cdot 2 = RHS \cdot 2$

{6 x + 8 y = 420 2 x + 3 y = 150

MultEqn

$(II):$ $LHS \cdot 3 = RHS \cdot 3$

{6 x + 8 y = 420 6 x + 9 y = 450

SysEqnSub

$(II):$ Subtract $(I)$

{6 x + 8 y = 420 6 x + 9 y - (6 x + 8 y) = 450 - 420

Distr

$(II):$ Distribute $- 1$

{6 x + 8 y = 420 6 x + 9 y - 6 x - 8 y = 450 - 420

SubTerms

$(II):$ Subtract terms

{6 x + 8 y = 420 y = 30

The

y -

coordinate of the point is

30 .

By substituting

30

for

y

into the equation, we can find the

x -

coordinate.

{6 x + 8 y = 420 y = 30

Substitute

$(I):$ $y = 30$

{6 x + 8 (30) = 420 y = 30

Multiply

$(I):$ Multiply

{6 x + 240 = 420 y = 30

SubEqn

$(I):$ $LHS - 240 = RHS - 240$

{6 x = 180 y = 30

DivEqn

$(I):$ $LHS / 6 = RHS / 6$

{x = 30 y = 30

Thus the fourth vertex is

(30, 30) .

\underline{Vertices} (0, 0) (0, 50) (30, 30) (70, 0)

d To find the number of each tree for the maximum

CO_{2}

absorption, we will substitute the vertices into the objective function. Thus, we can determine the maximum value of

A .

Let's begin with the vertex

(0, 0) .

A = 650 x + 300 y

SubstituteII

$x = 0$ , $y = 0$

A = 650 (0) + 300 (0)

ZeroPropMult

Zero Property of Multiplication

A = 0 + 0

AddTerms

Add terms

A = 0

For the vertex

(0, 0),

the absorption will be

0

lb/yr. We can find the

CO_{2}

absorption for the other vertices proceeding in the same way.

Vertex	Objective Function	Absorption (lb/yr)
$(0, 0)$	$A = 650 (0) + 300 (0)$	$0$
$(0, 50)$	$A = 650 (0) + 300 (50)$	$15000$
$(30, 30)$	$A = 650 (30) + 300 (30)$	$27500$
$(70, 0)$	$A = 650 (70) + 300 (0)$	$45500$

As a result, $70$ spruce trees should be planted to maximize the $CO_{2}$ absorption.

Subchapter links

Lesson Check p.160

Practice and Problem-Solving Exercises p.160-162

Standardized Test Prep p.162

Mixed Review p.162