Chapter ISG, Problem 8.7STP

a.

To determine

To find volume of small container.

Answer to Problem 8.7STP

Volume is $6x^2{y}^2-4x^{2}y-12x{y}^2+8xy$ unit³

Explanation of Solution

Given:

A cuboid.

Formula used: Volume of container, $v=length\times breadth\times height$

Calculation:

Length of container= $3y-2$

Breadth of container= $x-2$

Height of container= $2xy$

Substituting values,

  $v=(2xy)\times (x-2)\times (3y-2)$

Solving,

  $v=(2x^{2}y-4xy)\times (3y-2)$

So,

  $v=2x^{2}y(3y-2)-4xy(3y-2)$

And,

  $v=6x^2{y}^2-4x^{2}y-12x{y}^2+8xy$

Hence, volumeof container is $6x^2{y}^2-4x^{2}y-12x{y}^2+8xy$ unit³

b.

To determine

To find volume of large container

Answer to Problem 8.7STP

Volume of container is $15x^2{y}^2+5x{y}^2+60x^{3}y+20x^{2}y$

Explanation of Solution

Given:

A large container

Formula used:

Volume of container, $v=length\times breadth\times height$

Calculation:

Volume of container, $v=length\times breadth\times height$

Length of container= $y+4x$

Breadth of container= $5xy$

Height of container= $3x+1$

Substituting values,

  $v=(y+4x)\times (5xy)\times (3x+1)$

Solving,

  $v=(5x{y}^2+20x^{2}y)\times (3x+1)$

Now,

  $v=(3x+1)(5x{y}^2)+(3x+1)(20x^{2}y)$

Solving further,

  $v=15x^2{y}^2+5x{y}^2+60x^{3}y+20x^{2}y$

Hence, volume of container, $v=15x^2{y}^2+5x{y}^2+60x^{3}y+20x^{2}y$ .

c.

To determine

To find volume of water to be added

Answer to Problem 8.7STP

  $60x^{3}y+9x^2{y}^2+17x{y}^2+24x^{2}y-8xy$ is amount of water to be added

Explanation of Solution

Given:

Volumes of both container.

Calculation:

Volume of small container, $v_1^{}=6x^2{y}^2-4x^{2}y-12x{y}^2+8xy$

Volume of large container, $v_2^{}=15x^2{y}^2+5x{y}^2+60x^{3}y+20x^{2}y$

Water to be filled after emptying small container is= V₂ - V₁ Substituting values,

  =( $15x^2{y}^2+5x{y}^2+60x^{3}y+20x^{2}y$ )-( $6x^2{y}^2-4x^{2}y-12x{y}^2-8xy$ )

Solving, = $15x^2{y}^2+5x{y}^2+60x^{3}y+20x^{2}y$

  $-6x^2{y}^2+4x^{2}y+12x{y}^2-8xy$

Putting like terms together,

  = $60x^{3}y+(15x^2{y}^2-6x^2{y}^2)+(5x{y}^2+12x{y}^2)+20x^{2}y+4x^{2}y-8xy$

Adding,

  = $60x^{3}y+9x^2{y}^2+17x{y}^2+24x^{2}y-8xy$

Hence, amount of water to be added is $60x^{3}y+9x^2{y}^2+17x{y}^2+24x^{2}y-8xy$