Chapter 6.5, Problem 10E

(a)

To determine

Estimate the volume of pond using the Trapezoidal rule.

Answer to Problem 10E

Volume of pond,

  $\text{V}=26,360,000{\text{ft}}^{\text{3}}$

Explanation of Solution

Given information:

  

Vertical spacing is 200 ft.

Thus,

  $h=200\text{ft}$

According to Trapezoidal rule,

  $T=\frac{h}{2}\left(y_0+2y_1+2y_2+\mathrm{...}+2y_{n-1}+y_n\right)$

The average depth of pond,

  ${\text{D}}_{Avg}=20\text{ft}$

For volume of pond:

  $\begin{array}{l}\text{V}={\text{D}}_{Avg}\times T\\ =D_{Avg}\cdot \frac{h}{2}\left(y_0+2y_1+2y_2+\mathrm{...}+2y_{n-1}+y_n\right)\end{array}$

Substitute the values,

  $\begin{array}{l}\text{V}=\text{20}\cdot \frac{200}{2}\left[\begin{array}{l}0+2\left(520\right)+2\left(800\right)+2\left(1000\right)\\ +2\left(1140\right)+2\left(1160\right)+2\left(1110\right)+2\left(860\right)+0\end{array}\right]\\ =20\cdot 100\left(13180\right)\\ =26,360,000{\text{ft}}^{\text{3}}\end{array}$

Thus,

The volume of pond is 26,360,000 ft³.

(b)

To determine

Maximum number of licenses town can sell.

Answer to Problem 10E

Town can sell a maximum number of 988 licenses.

Explanation of Solution

Given information:

Season starts with one fish per 1000 cubic feet.

At the end of the season, 25% of the opening day’s fish population left.

Average seasonal catch is 20 fish per license.

From Part (a) result,

We have

The volume of pond is 26,360,000 ft³.

Now,

To get the number of fish in pond, divide the volume of pond by 1000, since one fish is allowed per 1000 cubic feet.

  $\text{Number}\text{of}\text{fish}\text{in}\text{pond}=\frac{26,360,000}{1,000}=26,360\text{fish}$

According to the statement,

25% of the fish will be left in the pond.

That means

75% of the fish will be caught.

  $\text{0}\text{.75}\left(26,360\right)=19,770\text{fish}$

Thus,

19,770 fish will be caught.

Now,

We have

The average seasonal catch is 20 fish per license.

Then

The maximum number of licenses available will be the number of fish allowed to be caught divided by the average seasonal catch per license.

Therefore,

  $\begin{array}{l}\text{Maximum}\text{number}\text{of}\text{licenses}=\frac{\text{Number}\text{of}\text{fish}\text{allowed}\text{to}\text{be}\text{caught}}{\text{Average}\text{seasonal}\text{per}\text{license}}\\ =\frac{19770}{20}\\ =988.5\\ \approx 988\text{licenses}\\ \end{array}$