Chapter 10, Problem 8CA

(a)

To determine

Find the perimeter.

Answer to Problem 8CA

  $56+8x$ in.

Explanation of Solution

Given Information: Length l = $16+2x$ in. and Width w = $12+2x$ in.

Formula Used : Perimeter of Rectangle = $2\times \left(Lenght+Width\right)$

Calculation:

Length l = $16+2x$ in. and Width w = $12+2x$ in.

Perimeter of table top = $2\times \left(Lenght+Width\right)$

So,

Perimeter of table top= $2\times \left(16+2x+12+2x\right)$ in.

Now,

Perimeter of tabletop = $2\times \left(28+4x\right)$ in.

Hence,

Perimeter of table top = $56+8x$ in.

(b)

To determine

Find the area.

Answer to Problem 8CA

  $4x^2+56x+192$ in².

Explanation of Solution

Given Information: Length l = $16+2x$ in. and Width w = $12+2x$ in.

Formula Used : Area of Rectangle = $Lenght\times Width$

Calculation:

Length l = $16+2x$ in. and Width w = $12+2x$ in.

Areaof table top = $Lenght\times Width$

So,

Area of table top = $\left(16+2x\right)\times \left(12+2x\right)$ in².

Now,

Area of table top = $192+32x+24x+4x^2$ in².

Hence,

Area of table top = $4x^2+56x+192$ in².

(c)

To determine

Find the possible values.

Answer to Problem 8CA

  $1,1.5,2,2.5$

Explanation of Solution

Given Information: Perimeter $<80$ in. and Area $\ge 252$ in².

Formula Used :Simplify

Calculation:

Perimeter $<80$ in. and Area $\ge 252$ in².

As,

Perimeter of table top = $56+8x$ in.

So,

  $56+8x<80$

Now,

  $8x<24$

So,

  $x<\frac{24}{8}$

Hence,

  $x<3$

As,

Area of table top = $4x^2+56x+192$ in².

So,

  $4x^2+56x+192\ge 252$

Now,

  $4x^2+56x\ge 252-192$

So,

  $4x\left(x+14\right)\ge 60$

Now,

  $x\left(x+14\right)\ge \frac{60}{4}$

So,

  $x\left(x+14\right)\ge 15$

Now,

  $x\ge 15$ or $x+14\ge 15$

So,

  $x\ge 15$ or $x\ge 1$

Hence,

Possible values of x are $1,1.5,2,2.5$