Modeling with Geometry

1. (a) Radius $= \frac{30}{2} = 15$ ft. Volume of cylinder:

$V = \pi r^2 h = \pi (15)^2 (50) = \pi (225)(50) = 11250\pi \approx 11250 \times 3.14159 \approx 35{,}343 \text{ ft}^3$

(b) Weight in pounds:

$35{,}343 \times 62.4 \approx 2{,}205{,}403 \text{ lb}$

Weight in tons:

$\frac{2{,}205{,}403}{2000} \approx 1{,}103 \text{ tons}$

2. (a) Volume of cylinder ($r = 8$, $h = 24$):

$V_{\text{cyl}} = \pi (8)^2 (24) = 1536\pi \approx 4{,}825 \text{ ft}^3$

(b) Volume of hemisphere (half of $\frac{4}{3}\pi r^3$):

$V_{\text{hemi}} = \frac{1}{2} \cdot \frac{4}{3}\pi (8)^3 = \frac{2}{3}\pi (512) = \frac{1024\pi}{3} \approx 1{,}072 \text{ ft}^3$

(c) Total volume:

$V_{\text{total}} = 4{,}825 + 1{,}072 \approx 5{,}897 \text{ ft}^3$

3. (a) Area of city block:

$A = 250 \times 120 = 30{,}000 \text{ ft}^2$

(b) Estimated population:

$30{,}000 \times 0.008 = 240 \text{ people}$

(c) Average residents per floor:

$\frac{240}{15} = 16 \text{ people per floor}$

4. (a) Area of park:

$A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(90)(60) = 2{,}700 \text{ m}^2$

(b) Perimeter:

$P = 75 + 80 + 90 = 245 \text{ meters}$

(c) Number of benches at every $15$ meters:

$\frac{245}{15} \approx 16.3$

Since benches must be whole units, $\lfloor 16.3 \rfloor = 16$ gaps of $15$ m means $16$ benches can be placed (assuming one at each interval marker, not at the very start). However, if one bench is placed at each $15$-meter mark including the start, there are $17$ positions. The most natural interpretation: 16 benches are needed (one every 15 m covers the 245 m path with the last interval being shorter than 15 m).

5. (a) Outer volume ($r_{\text{outer}} = 4.5$ ft):

$V_{\text{outer}} = \frac{4}{3}\pi (4.5)^3 = \frac{4}{3}\pi (91.125) = 121.5\pi \approx 381.7 \text{ ft}^3$

(b) Inner radius $= 4.5 - 0.25 = 4.25$ ft:

$V_{\text{inner}} = \frac{4}{3}\pi (4.25)^3 = \frac{4}{3}\pi (76.766) \approx 102.354\pi \approx 321.6 \text{ ft}^3$

(c) Percentage usable:

$\frac{321.6}{381.7} \times 100 \approx 84.2\%$

6. (a) Usable roof dimensions (subtract $1$ ft border on each side):

• Length: $40 - 2(1) = 38$ ft
• Width: $25 - 2(1) = 23$ ft

$A_{\text{usable}} = 38 \times 23 = 874 \text{ ft}^2$

(b) Each panel is $5.5$ ft by $3.25$ ft. Find how many fit in each direction:

Orientation 1 (5.5 ft along length, 3.25 ft along width):

• Along length: $\lfloor 38 \div 5.5 \rfloor = \lfloor 6.909 \rfloor = 6$ panels
• Along width: $\lfloor 23 \div 3.25 \rfloor = \lfloor 7.077 \rfloor = 7$ panels
• Total: $6 \times 7 = 42$ panels

Orientation 2 (3.25 ft along length, 5.5 ft along width):

• Along length: $\lfloor 38 \div 3.25 \rfloor = \lfloor 11.69 \rfloor = 11$ panels
• Along width: $\lfloor 23 \div 5.5 \rfloor = \lfloor 4.18 \rfloor = 4$ panels
• Total: $11 \times 4 = 44$ panels

The best orientation gives 44 panels.

(c) Area covered by panels:

$44 \times (5.5 \times 3.25) = 44 \times 17.875 = 786.5 \text{ ft}^2$

Total roof area: $40 \times 25 = 1000 \text{ ft}^2$

$\frac{786.5}{1000} \times 100 = 78.7\%$