The surface area of the curved surface of the cone is given by the - - PowerPoint PPT Presentation

DAY 166 – CONICAL

STRUCTURES

INTRODUCTION

Quite a number of structures made in the form of a cone. These include tents, conical ice cream scoop, and some roofs. When designing these structures, one may want it to have the maximum volume with the available size of the material. One may also want a conical structure with a given volume to have the minimum surface area to minimize the cost of the material. In this lesson, we will apply geometric methods to solve design problems of the conical structures or objects to satisfy physical constraints

• r minimize cost.

VOCABULARY

Cone This a three dimensional solid with a circular base and a curved surface that results to a single apex.

A cone with a base radius 𝑠, a vertical height ℎ, and a slant height 𝑚 has a volume

1 3 𝜌𝑠2 × ℎ.

The surface area of the curved surface of the cone is given by the formula 𝜌𝑠𝑚. In case of a closed circular part of the cone, the total surface area of the cone will be 𝜌𝑠2 + 𝜌𝑠𝑚. Since by Pythagorean theorem the slant height 𝑚 = ℎ2 + 𝑠2, the surface area of a cone can be given by 𝜌𝑠2 + 𝜌𝑠 ℎ2 + 𝑠2. We use these formulas in solving design problems of the conical structures or objects to satisfy physical constraints or minimize cost.

Lets consider a canvas with a surface area of 50 𝑔𝑢2 which we intend to use in making a conical tent with a maximum possible volume. Let r and h be the radius and height of the conical tent respectively. Volume, V =

1 3 𝜌𝑠2 ℎ

Surface area = 50 = 𝜌𝑠 ℎ2 + 𝑠2 Squaring both sides and making h the subject of the formula we get, ℎ =

2500 𝜌2𝑠2 − 𝑠2

1 2

Substituting h in the volume formula we get, 𝑊 =

1 3 𝜌𝑠2 2500 𝜌2𝑠2 − 𝑠2

1 2

Taking 𝜌 = 3.142, the function to be maximized is 𝑊(𝑠) =

1 3 × 3.142 × 𝑠2 2500 3.1422×𝑠2 − 𝑠2

1 2

𝑊(𝑠) = 1.047𝑠2

253.2 𝑠2 − 𝑠2

1 2

Let’s assume that we want to make a closed cone which will hold 10 𝑔𝑢3 of water with minimum possible surface area to reduce the cost of the material. Surface area, 𝑇 = 𝜌𝑠2 + 𝜌𝑠 ℎ2 + 𝑠2

1 2

Volume V, = 10 𝑔𝑢3 =

1 3 𝜌𝑠2ℎ ⟹ ℎ = 30 𝜌𝑠2

Substituting the value of ℎ we get, 𝑇 = 𝜌𝑠2 + 𝜌𝑠

900 𝜌2𝑠4 + 𝑠2

1 2

Taking 𝜌 = 3.142, we get the function, 𝑇(𝑠) = 3.142𝑠2 + 3.142𝑠

91.17 𝑠4 + 𝑠2

1 2

We then find the minimum value of this function to get the minimum surface area that can hold 10 𝑔𝑢3.

Example A man wants to make a conical porous container to hold 2 𝑔𝑢3 of water. He wants the container to have the maximum surface area to cool the water faster. Write the surface area as a function of the radius of the container.

Solution Surface area, 𝑇 = 𝜌𝑠2 + 𝜌𝑠 ℎ2 + 𝑠2

1 2

Volume V, = 2 𝑔𝑢3 =

1 3 𝜌𝑠2ℎ ⟹ ℎ = 6 𝜌𝑠2

Substituting the value of ℎ we get, 𝑇 = 𝜌𝑠2 + 𝜌𝑠

36 𝜌2𝑠4 + 𝑠2

1 2

Taking 𝜌 = 3.142, we get the function, 𝑇(𝑠) = 3.142𝑠2 + 3.142𝑠

3.647 𝑠4 + 𝑠2

1 2

HOMEWORK A canvas has a surface area of 250 𝑔𝑢3. If this canvas is used to make a conical tent, write the volume of the tent as a function of its radius.

ANSWERS TO HOMEWORK

𝑊(𝑠) = 1.047𝑠2

6331 𝑠2 − 𝑠2

1 2

THE END