Chapter 3.8, Problem 12CFU

a.

To determine

To find: an equation that represents the given situation.

Answer to Problem 12CFU

  $V=\frac{16}{5}h{g}^2$

Explanation of Solution

Given information:

Let the volume of wood be $V$ , height be $h$ and grith be $g$ .

The variable V varies jointly as h and $g^2$ .

  $\begin{array}{l}h=40m\\ g=1.5m\\ V=288m^3\end{array}$

Calculation:

The equation for the given condition will be as follows:

  $V=kh{g}^2$...........(1)

Put the values of $V,h$ and $g$ in the above equation to evaluate the constant of variation $\left(k\right)$ as follows:

  $\begin{array}{c}288=k\left(40\right){\left(1.5\right)}^2\\ k=\frac{288}{\left(40\right){\left(1.5\right)}^2}\\ =\frac{16}{5}\end{array}$

Put the value of k in equation (1) to get,

  $V=\frac{16}{5}h{g}^2$

Thus, the equation representing the given situation is

  $V=\frac{16}{5}h{g}^2$.

b.

To determine

To find: the volume of wood.

Answer to Problem 12CFU

  $24000m^3$

Explanation of Solution

Given information:

Let the volume of wood be $V$ , height be $h$ , grith be $g$ and number of trees be n.

  $\begin{array}{l}h=75m\\ g=2m\\ V=?\\ n=50\end{array}$

The equation evaluated is as follows:

  $V=\frac{16}{5}h{g}^2$

Calculation:

Put the values in the given equation to evaluate the volume of one tree as shown below.

  $\begin{array}{c}V=\frac{16}{5}h{g}^2\\ =\frac{16}{5}\left(75m\right){\left(2m\right)}^2\\ =480m^3\end{array}$

Since the volume of one tree is $480m^3$ . The volume of 50 trees can be evaluated as follows:

  $\begin{array}{c}{V}_{50trees}=nV\\ =50\left(480m^3\right)\\ =24000m^3\end{array}$

Thus, the volume of 50 trees is $24000m^3$ .