The Goodsmell perfume producing company has a new line of perfume and is designing a new bottle for it. Because of the expense of the glass required to make the bottle, the surface area must be less than 150 cm2. The company also wants the bottle to hold at least 100mL of perfume. The design under consideration is in the shape of a cylinder. Determine the maximum volume possible for a cylindrical bottle that has a total surface area of less than 150 cm2. Determine the volume to the nearest 10mL. Report the dimensions of the bottle and the corresponding surface area and volume. This cylindrical perfume bottle needs to be boxed in a prism for sale on store shelves. The Goodsmell perfume producing company would like a box with the smallest possible surface area which will hold your design for the perfume bottle. Report the dimensions of the box and the corresponding area and volume. The pretty perfume company, Goodsmell’s competition, has also designed a new perfume bottle. The bottle is to be a spherical shape with a diameter of 7cm. Determine the volume and surface area of this bottle. The spherical bottle has a conical shaped lid with a diameter of 5cm and a height of 4.5cm. Given this information, calculate the volume and surface area of the lid of the spherical shaped bottle. Determine the dimensions of the smallest possible box which could be used to package Pretty Perfume’s new bottle with its spherical bottle and conical-shaped lid. However, this box is not shaped like a rectangular prism. Pretty Perfume would like to have unique packaging and have chosen to have a box shaped like a pyramid. Calculate the volume and surface area of this pyramid shaped box. The Final Product: Prepare a written report that includes: A clear and concise explanation of the process that you used to solve the problem. The calculations that you made, presented in an organized fashion. A rationale (reason) for your selection of values.

1. r = 2.82 cm

h = 5.6 cm

The maximum volume possible to the nearest 10 mL = 140 mL

2. Size side of square base of box is 5.64 cm

Height of box = 5.6 cm

The surface area of the box is 189.96 cm²

The volume of the box is 178.13 cm³

3.  The procedure for solving the problem was through noting that the shape of the cross-section of the pyramid is an isosceles triangle ans also that smallest possible box for the pretty perfume is one which fits the angle of inclination of the lid. This was found out by initially using the combined height of the perfume and the lid (placed to fit the spherical outline of the bottle) to calculate the dimensions of the pyramid, from which it was observed that the angle of inclination of the lid is larger than that of the calculated dimension, such that the lid outline would be visible and could eventually tear the perfume box  

With the inclination angle, β, which is the base angle of the isosceles triangle, the angle at the top of the pyramid cross-section is calculated and the following relations are used to calculate the triangular cross-section of the pyramid

h = a·cos(α/2)

b = 2·a·cos(β)

With the calculated dimensions, a, b, and h the area, A, of the square pyramid is calculated as 2×b×a + b² and the volume, V, as 1/3×b²×h

The attached diagram shows the the cross-section of the perfume in the pyramid box.

Step-by-step explanation:

1. The surface area of the cylinder = 2πr² + 2πrh = 150 cm².........(1)

The volume of the cylinder, V = πr²h = 100 mL = 100 cm³..............(2)

From equation (2), h = 100/(π·r²)

Substituting the value if h in equation (1), we have;

2πr² + 2πr100/(π·r²) = 150

2πr² + 200/r = 150

(2πr³ + 200)/r = 150

2πr³ + 200 = 150×r

2πr³ -150·r+ 200 = 0

150 = 2πr² + 2πrh

h = (150 - 2πr²)/(2πr)

h = (75- πr²)/(πr)

Substituting the value of h in the equation for the volume, we have;

V = πr²h =  πr²(75- πr²)/(πr)

V = 75·r - π·r³

At maximum volume, dV/dr = 0, we have

d(75·r - π·r³)/dr = 75 - 3·π·r²= 0

3·π·r²= 75

π·r² = 25

r = 5√π/π

h = (75- πr²)/(πr) = (75- π(5√π/π)²)/(π(5√π/π)) = (75 -25)/(5·√π)

h = 50/(5·√π)= 10·√π/π

The maximum volume = πr²h = π×25/π×10·√π/π =  250·√π/π = 141.05 cm³

The maximum volume possible = 141.05 cm³ = 141.05 mL

The maximum volume possible to the nearest 10 mL = 140 mL

The dimensions of the bottle are;

r = 2.82 cm

h = 5.6 cm

The surface area of the bottle = 2π(2.82)² + 2π×2.82 ×5.6 = 149.2 cm

2

Given that the cylindrical bottle has r = 2.82 cm and h = 5.6 cm, we have;

Size side of square base of box = 2 × 2.82 = 5.64 cm

Height of box = 5.6 cm

The surface area of the box = 2 × Area of base + 4 × Area of side

The surface area of the box = 2 *5.64^2 + 4 * 5.6 * 5.64 = 189.96 cm²

The volume of the box = Area of base × Height = 5.64^2*5.6 = 178.13 cm³

3. Diameter of spherical bottle = 7 cm = 2×r

Volume of the sphere  bottle = 4/3πr³ = 4/3*3.5^3*π = 343/6·π = 179.6 cm³

The surface area of the sphere  bottle = 4πr² = 4*(7/2)^2*π = 49·π = 156.94 cm²

3 i. The volume of a cone = 1/3πr²h = 1/3*(5/2)^2*4.5 = 9.385·π = 29.45 cm³

The  surface area of a cone = πrS

S = √(4.5^2 + (5/2)^2) = 5.15

The  surface area of a cone = π*2.5*5.15 = 40.43 cm²

3 ii. The depth of fitness of the lid on the bottle = 7/2 - √(7/2)^2 - 2.5^2) = 1.05

The total height of the spherical bottle with the conical lid = 7 + 4.5 - 1.05 = 10.45 cm

3 iii. Given that the box is shaped like a pyramid we have;

Width of the box at middle of the height of the spherical bottle = 7 cm

Height of the box = 10.45 cm

With the aid of a graphing calculator, the width of the square pyramid is found to be 12.12 cm

The volume = 1/3*12.12^2*10.45 = 511.68 cm²

The surface area = 2*12.12*√(12.12/2)^2 + 10.45^2) +12.12²= 439.7 cm²

The angle of inclination of the lid = tan⁻¹ (4.5/2.5) = 60.95°

The angle of inclination of the calculated box is tan⁻¹ (10.45/6.06) = 59.88

Since the lid is steeper, we make use of the angle of the lid

The base angles are thus = 60.95°

The angle at the top is thus 180 - 60.95*2 = 58.11°

Therefore, by the formula, we find that

a = 12.25 cm

b = 11.897 cm

h = a·cos(α/2)

h = 10.707 cm

The volume = 1/3*11.897^2*10.707 = 505.15 cm³

The surface area = 2*11.897*√(11.897/2)^2 + 10.707^2) +11.897²= 432.98 cm²

The angle at the top of the box = 2

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