Exercise 8 Page 581

Practice makes perfect

a We are asked to write a polynomial in standard form that represents the perimeter of the tabletop. Let's take a look at the given diagram.

We will use the formula for the perimeter of a rectangle.

P = 2 ℓ + 2 w

Here,

w

is the width of the rectangle and

ℓ

is its length. We can see from the diagram that the width

w

of the rectangle consists of the width of the tiled part and two widths of the mosaic border

x .

Similarly, the length of the rectangle $ℓ$ is the length of the tiled part plus two widths of the mosaic border $x .$

To write the polynomial representing the perimeter of our tabletop, we can substitute the values of

ℓ

and

w

into the formula.

P = 2 ℓ + 2 w ⇓ P = 2 (16 + 2 x) + 2 (12 + 2 x)

For our polynomial to be in standard form, we need to distribute the coefficients to remove the parentheses and the exponents of the terms must decrease from left to right. Let's rewrite our expression in this form!

P = 2 (16 + 2 x) + 2 (12 + 2 x)

▼

Simplify right-hand side

Distr

Distribute $2$

P = 32 + 4 x + 24 + 4 x

AddTerms

Add terms

P = 56 + 8 x

CommutativePropAdd

Commutative Property of Addition

P = 8 x + 56

b Now we want to write a polynomial in standard form that represents the area of the tabletop. To do so, we will recall the formula for the area of a rectangle.

A = ℓ w

In Part A, we found that

ℓ = 16 + 2 x

and

w = 12 + 2 x .

To obtain a polynomial, we can substitute these values into the formula.

A = (16 + 2 x) (12 + 2 x)

Now, let's rewrite this expression in standard form!

A = (16 + 2 x) (12 + 2 x)

▼

Multiply parentheses

Distr

Distribute $(16 + 2 x)$

A = (16 + 2 x) 12 + (16 + 2 x) 2 x

Distr

Distribute $12 & 2 x$

A = 192 + 24 x + 32 x + 4 x^{2}

AddTerms

Add terms

A = 192 + 56 x + 4 x^{2}

CommutativePropAdd

Commutative Property of Addition

A = 4 x^{2} + 56 x + 192

c Finally, we want to select all of the possible values of

x

from the given list which would make the tabletop's perimeter less than

80

inches while ensuring that its area is at least

252

square inches. We can represent this information as a system of inequalities.

{P < 80 A \geq 252 (I) (II)

Now, we will express

P

and

A

in terms of

x

by using the polynomials we found in Parts A and B.

{8 x + 56 < 80 4 x^{2} + 56 x + 192 \geq 252 (I) (II)

We want to find the values of

x

that satisfy these inequalities. Let's start by simplifying Inequality (I).

8 x + 56 < 80

SubIneq

$LHS - 56 < RHS - 56$

8 x < 24

DivIneq

$LHS / 8 < RHS / 8$

x < 3

We know that for the tabletop's perimeter to be less than

80

inches, the value of

x

must be less than

3 .

Let's remove these values from the list of possible values for

x .

Let's now focus on Inequality (II). First, we will gather all the terms on one side and simplify.

4 x^{2} + 56 x + 192 \geq 252

SubIneq

$LHS - 252 \geq RHS - 252$

4 x^{2} + 56 x - 60 \geq 0

DivIneq

$LHS / 4 \geq RHS / 4$

x^{2} + 14 x - 15 \geq 0

We want to check which of the remaining values of

x

will satisfy this inequality. To do so, we can substitute them into the inequality.

$x$	$x^{2} + 14 x - 15$	$x^{2} + 14 x - 15 \geq 0$
$0.5$	$0.5^{2} + 14 (0.5) - 15$	$- 7.75 ≱ 0$
$1$	$1^{2} + 14 (1) - 15$	$0 \geq 0$
$1.5$	$1.5^{2} + 14 (1.5) - 15$	$8.25 \geq 0$
$2$	$2^{2} + 14 (2) - 15$	$17 \geq 0$
$2.5$	$2.5^{2} + 14 (2.5) - 15$	$26.25 \geq 0$

We can see that only $1,$ $1.5,$ $2,$ and $2.5$ make the perimeter smaller than $80$ inches and the area at least $252$ square inches.

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Exercise 8 Page 581

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