We were told that Arianna needed

140 x^{2} + 60 x

square inches of paper to make a book jacket

10 x

inches tall. Since the paper cover has a rectangular shape when unfolded, its area is given by its height

h

times its length

ℓ .

Using this information, we can solve solve for the length.

A = h ℓ

▼

Solve for

h

$LHS / ℓ = RHS / ℓ$

\frac{A}{ℓ} = h

Rearrange equation

h = \frac{A}{ℓ}

Therefore, since we know that

A = 140 x^{2} + 60 x

and

h = 10 x,

we can calculate the quotient of the polynomials

\frac{1 4 0 x ^{2} + 6 0 x}{1 0 x}

to find the length of the paper cover.

\frac{1 4 0 x ^{2} + 6 0 x}{1 0 x}

▼

Simplify

Write as a sum of fractions

\frac{1 4 0 x ^{2}}{1 0 x} + \frac{6 0 x}{1 0 x}

$\frac{a ^{m}}{a ^{n}} = a^{m - n}$

\frac{1 4 0}{1 0} x + \frac{6 0}{1 0} x^{0}

$a^{0} = 1$

\frac{1 4 0}{1 0} x + \frac{6 0}{1 0}

Calculate quotient

14 x + 6

Now we know that the length of the book cover is

14 x + 6

inches. We are also told that the spine is

2 x

inches wide, the front and the back are

6 x

inches wide, and that she allowed for a front and back flap. Let's take look at the diagram for the paper cover.

In the diagram shown

f

represents the flap length. We can see that the paper cover length is

f + 6 x + 2 x + 6 x + f = 14 x + 2 f .

We can equalize this to the length we calculated before

14 x + 6

and solve for

f .

14 x + 2 f = 14 x + 6

▼

Solve for

f

$LHS - 14 x = RHS - 14 x$

2 f = 6

$LHS / 2 = RHS / 2$

f = 6

Therefore, the length for each flap is

6

inches. Knowing this, we have the dimensions for all the parts of the paper cover and we can make the book jacket.

Exercise 48 Page 316

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