Chapter ISG, Problem 8.7.1.1P

(a)

To determine

To find: The dimension of small, medium and large boxes.

Answer to Problem 8.7.1.1P

Dimension of extra-large boxes: $\left(n+10\right)\left(n+10\right)\left(n-10\right)$

Volume of extra-large boxes: $n^3+10n^2-100n-1000{\text{ in}}^3$

Dimension of small boxes: $\left(n+4\right)\times \left(n+4\right)\times \left(n-4\right)$

Dimension of medium boxes: $\left(n+6\right)\times \left(n+6\right)\times \left(n-6\right)$

Dimension of large boxes: $\left(n+8\right)\times \left(n+8\right)\times \left(n-8\right)$

Explanation of Solution

Given: The volume of small boxes is represented by $n^3+4n^2-16n-64{\text{ in}}^3$

The volume of medium boxes is represented by $n^3+6n^2-36n-216{\text{ in}}^3$

The volume of large boxes is represented by $n^3+8n^2-64n-512{\text{ in}}^3$

For small boxes: $n^3+4n^2-16n-64{\text{ in}}^3$

Factor the cubic polynomial: $\left(n+4\right)\left(n+4\right)\left(n-4\right)$

Dimension of small boxes: $\left(n+4\right)\times \left(n+4\right)\times \left(n-4\right)$

For medium boxes: $n^3+6n^2-36n-216{\text{ in}}^3$

Factor the cubic polynomial: $\left(n+6\right)\left(n+6\right)\left(n-6\right)$

Dimension of medium boxes: $\left(n+6\right)\times \left(n+6\right)\times \left(n-6\right)$

For Large boxes: $n^3+8n^2-64n-512{\text{ in}}^3$

Factor the cubic polynomial: $\left(n+8\right)\left(n+8\right)\left(n-8\right)$

Dimension of large boxes: $\left(n+8\right)\times \left(n+8\right)\times \left(n-8\right)$

Using above patter the dimension of extra-large boxes: $\left(n+10\right)\left(n+10\right)\left(n-10\right)$

Volume of extra-large boxes: $n^3+10n^2-100n-1000{\text{ in}}^3$

(b)

To determine

To find: The area of small, medium and large boxes.

Answer to Problem 8.7.1.1P

Area of small boxes: $2\left(3n^2+8n-16\right)$

Area of medium boxes: $2\left(3n^2+12n-36\right)$

Area of large boxes: $2\left(3n^2+16n-64\right)$

Area of extra-large boxes is $2\left(3n^2+20n-100\right)$

Explanation of Solution

Given: The volume of small boxes is represented by $n^3+4n^2-16n-64{\text{ in}}^3$

The volume of medium boxes is represented by $n^3+6n^2-36n-216{\text{ in}}^3$

The volume of large boxes is represented by $n^3+8n^2-64n-512{\text{ in}}^3$

Dimension of small boxes: $\left(n+4\right)\times \left(n+4\right)\times \left(n-4\right)$

Area of small box: $2\left[\left(n+4\right)\left(n+4\right)+\left(n+4\right)\left(n-4\right)+\left(n+4\right)\left(n-4\right)\right]$

Area $=2\left(3n^2+8n-16\right)$

Dimension of medium boxes: $\left(n+6\right)\times \left(n+6\right)\times \left(n-6\right)$

Area = $2\left[\left(n+6\right)\left(n+6\right)+\left(n+6\right)\left(n-6\right)+\left(n+6\right)\left(n-6\right)\right]=2\left(3n^2+12n-36\right)$

Dimension of large boxes: $\left(n+8\right)\times \left(n+8\right)\times \left(n-8\right)$

Area = $2\left[\left(n+8\right)\left(n+8\right)+\left(n+8\right)\left(n-8\right)+\left(n+8\right)\left(n-8\right)\right]=2\left(3n^2+16n-64\right)$

Using patter of area of boxes.

Area of extra-large boxes is $2\left(3n^2+20n-100\right)$

(c)

To determine

To find: The volume of extra large boxes.

Answer to Problem 8.7.1.1P

Volume of extra-large boxes: $n^3+10n^2-100n-1000{\text{ in}}^3$

Explanation of Solution

Given: The volume of small boxes is represented by $n^3+4n^2-16n-64{\text{ in}}^3$

The volume of medium boxes is represented by $n^3+6n^2-36n-216{\text{ in}}^3$

The volume of large boxes is represented by $n^3+8n^2-64n-512{\text{ in}}^3$

Area of extra-large boxes is $2\left(3n^2+20n-100\right)$ .

Volume of box is length times width times height.

Area = $2\left[\left(n+10\right)\left(n+10\right)+\left(n+10\right)\left(n-10\right)+\left(n+10\right)\left(n-10\right)\right]$

So, volume is $n^3+10n^2-100n-1000{\text{ in}}^3$