Chapter 7.1, Problem 63E

a.

To determine

The graph of the two log rules by using graphing utility.

Answer to Problem 63E

  $(24.72,429.3)$ .

Explanation of Solution

Given information:

For a $16$ -foot log, the Doyle Log Rule is modelled by $V_1={(D-4)}^2,5\le D\le 40$ ,and the Scribner Log Rule is modelled by $V_2=0.79D^2-2D-4,5\le D\le 40$ ,where $D$ is the diameter(in inches) of the log and $V$ is its volume(in board feet).

  $(a)$ Use a graphing utility to graph the two log rules in the same viewing window.

Calculation:

The given information states that a $16$ -foot log, the Doyle Log Rule is modelled by $V_1={(D-4)}^2,5\le D\le 40$ ,and the Scribner Log Rule is modelled by $V_2=0.79D^2-2D-4,5\le D\le 40$ ,where $D$ is the diameter(in inches) of the log and $V$ is its volume(in board feet).

Byusing a graphing utility to graph the two log rules in the same viewing window are as shown below:

  

Thus, both the solution of equations are intersected at point $(24.72,429.3)$ .

b.

To determine

The diameter of the two log rules.

Answer to Problem 63E

  $x\approx 24.72$ .

Explanation of Solution

Given information:

For a $16$ -foot log, the Doyle Log Rule is modelled by $V_1={(D-4)}^2,5\le D\le 40$ ,and the Scribner Log Rule is modelled by $V_2=0.79D^2-2D-4,5\le D\le 40$ ,where $D$ is the diameter(in inches) of the log and $V$ is its volume(in board feet).

  $(b)$ For what diameter do the two rules agree?

Calculation:

The given information states that a $16$ -foot log, the Doyle Log Rule is modelled by $V_1={(D-4)}^2,5\le D\le 40$ ,and the Scribner Log Rule is modelled by $V_2=0.79D^2-2D-4,5\le D\le 40$ ,where $D$ is the diameter(in inches) of the log and $V$ is its volume(in board feet).

Byusing a graphing utility find the diameter of the two rules are as shown below:

  

Thus, the diameter is $x\approx 24.72$ .

c.

To determine

To explain:The reasoning from the given statement.

Answer to Problem 63E

Doyle rule.

Explanation of Solution

Given information:

For a $16$ -foot log, the Doyle Log Rule is modelled by $V_1={(D-4)}^2,5\le D\le 40$ ,and the Scribner Log Rule is modelled by $V_2=0.79D^2-2D-4,5\le D\le 40$ ,where $D$ is the diameter(in inches) of the log and $V$ is its volume(in board feet).

  $(c)$ You are selling large logs by the board foot. Which rule would you use? Explain your reasoning.

Calculation:

For large logs diameter is approximately $25$ in and more, the Doyle Rule calculates greater volume.

The more volume sell by Doyle rule.