1. ## HELP with function

Hey can some one pls explain this problem to me! tnx in advance

The swimming pool shown in Figure 1.54 is 3 ft. deep at the shallow end, 8 ft. deep at the deepest end, 40 ft. long, 30 ft. wide and the bottom is an incline plane. Express the volume V of water in the pool as a function of the height h of the water above the deep end.(hint: V will be a piecewise define function)

2. Originally Posted by geeko
Hey can some one pls explain this problem to me! tnx in advance

The swimming pool shown in Figure 1.54 is 3 ft. deep at the shallow end, 8 ft. deep at the deepest end, 40 ft. long, 30 ft. wide and the bottom is an incline plane. Express the volume V of water in the pool as a function of the height h of the water above the deep end.(hint: V will be a piecewise define function)
from your diagram, are you aware that you can split the volume of the figure up into the volume of a box and a triangular prism? can you find the volume of each of those figures?

3. Hello, geeko!

A swimming pool is 3 ft. deep at the shallow end, 8 ft. deep at the deep end,
40 ft. long, 30 ft. wide, and the bottom is an incline plane.
Express the volume $V$ of water in the pool as a function of the height $h$
of the water at the deep end. (Hint: $V$ will be a piecewise-defined function)
Code:
                       40
-   * - - - - - - - - - - - - - - - - - *
:   |                                   |
3   |                                   | 3
:   |                   40              |
-   * - - - - - - - - - - - - - - - - - *
:   |          x                  *
:   + - - - - - - - - - - - *
6   |:::::::::::::::::*
:  h|:::::::::::*
:   |:::::*
-   *

For the first six feet, the volume of water is a triangular prism.
. . Its volume is: . $\text{Area of triangle} \times \text{30-foot width}$

From the similar right triangles: . $\frac{x}{h} = \frac{40}{6} \quad\Rightarrow\quad x \:=\:\frac{20}{3}h$ .[1]

The area of the triangle is: . $A \:=\:\frac{1}{2}hx$

Substitute [1]: . $A \:=\:\frac{1}{2}h\left(\frac{20}{3}h\right) \:=\;\frac{10}{3}h^2$

Then the volume is: . $V \:=\:\frac{10}{3}h^2 \times 30 \:=\:100h^2$

. . $\boxed{\text{For }0 \leq h \leq 6\!:\;\;V \;=\;100h^2\text{ ft}^3}$

For the last three feet of water, the bottom (triangular) portion is already filled.
. . It contains: . $100(6^2) \:=\:3600$ ft³ of water.

The upper portion is a rectangular slab of water:
. . length 40, width 30, and height $h-6$
Its volume is: . $40 \times 30 \times (h-6) \:=\:1200(h-6)$ ft³.

The total volume is: . $V \:=\:1200(h-6) + 3600$

. . $\boxed{\text{For }6 \leq h \leq 9\!:\;\;V \;=\;1200(h-3)\text{ ft}^3}$

Therefore: . $V \;=\;\bigg\{ \begin{array}{ccc}100h^2 & & 0 \leq h \leq 6 \\ 1200(h-3) && 6 \leq h \leq 9 \end{array}$