Calculus 1: MTH 150 Quiz 7 Solution - Optimization and Cylinders

This document presents the complete solutions for MTH 150 Calculus 1 Quiz 7, addressing two key problems. The first problem focuses on determining the optimal height of a lamppost to maximize illumination over a circular pathway, utilizing trigonometric principles and optimization techniques to find the angle that maximizes light intensity. The second problem involves designing a cylindrical soda can, specifically minimizing its surface area while maintaining a fixed volume. This solution demonstrates the application of calculus, including finding derivatives and solving for critical points, to optimize the dimensions of the cylinder and provides the calculated height and radius. The solution also outlines the assumptions made during the calculations. The provided solutions offer a clear, step-by-step approach to solving these calculus problems.

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1. As from the figure:
𝑠 = 20
cos 𝜃
𝐼 = sin(𝜃)
20
cos(𝜃)
=
sin(𝜃) cos(𝜃)
20
sin(𝜃) cos(𝜃) = 1
2sin(2𝜃)
∴ 𝐼 =sin(2𝜃)
40
For maximum intensity 𝑑𝐼
𝑑𝜃 = 0
⟹ 𝑑𝐼
𝑑𝜃= cos(2𝜃)
20 = 0
cos(2𝜃) = 0
𝜃 =𝜋
4 + 𝜋𝑛 𝑂𝑅 𝜃 =
3𝜋
4 + 𝜋𝑛 𝑓𝑜𝑟 𝑛 = 0, 1, 2, 3 …
𝑓𝑜𝑟 𝑛 = 0, 𝜃 =
𝜋
4 𝑜𝑟 3𝜋
4
𝑇ℎ𝑢𝑠 𝜃 =
𝜋
4 𝑟𝑎𝑑 𝑂𝑟 45 °
tan(𝜃) = 𝐻𝑒𝑖𝑔ℎ𝑡
20
𝐻𝑒𝑖𝑔ℎ𝑡 = 20 tan(45°)
= 20 𝑓𝑡
2. A cylindrical soda can is being designed to hold a volume of approximately 21.7 in3.

a) Minimize surface area
Since we know the volume we can solve for h in terms of r
𝑉 = 𝜋𝑟2ℎ
∴ ℎ =21.7
𝜋𝑟2
𝑆 = 2𝜋𝑟2 + 2𝜋𝑟ℎ
𝑆 = 2 (𝜋𝑟2 + 𝜋𝑟 (
21.7
𝜋𝑟2 )) = 2𝜋𝑟2 + 43.4𝑟−1
For minimum r, 𝑑𝑆
𝑑𝑟 = 0
𝑑𝑆
𝑑𝑟= 4𝜋𝑟 −
43.4
𝑟2 = 0
4𝜋𝑟3 = 43.4
𝑟 = √43.4
4𝜋
3
= 𝟏. 𝟓𝟏𝟏𝟓𝟔𝟒𝟐𝟖𝟓 𝒊𝒏
∴ ℎ = 21.7
𝜋(1.511564)2 = 𝟑. 𝟎𝟐𝟑𝟏𝟐𝟖𝟓𝟔𝟗 𝒊𝒏
b) The assumption that my read to that error are:
i) The cylindrical soda is a closed cylinder
ii) The surface area function is differentiable for all values of r