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

D AY 166 – C ONICAL STRUCTURES

I NTRODUCTION 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 or minimize cost.

V OCABULARY 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 1 3 𝜌𝑠 2 × ℎ. a slant height 𝑚 has a volume 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 𝜌𝑠 2 + 𝜌𝑠𝑚. Since by Pythagorean theorem the slant height ℎ 2 + 𝑠 2 , the surface area of a cone can be given 𝑚 = by 𝜌𝑠 2 + 𝜌𝑠 ℎ 2 + 𝑠 2 . 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 𝑔𝑢 2 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. 1 3 𝜌𝑠 2 ℎ Volume, V = S urface area = 50 = 𝜌𝑠 ℎ 2 + 𝑠 2 Squaring both sides and making h the subject of the 1 2500 formula we get, ℎ = 2 𝜌 2 𝑠 2 − 𝑠 2 Substituting h in the volume formula we get, 1 1 2500 2 3 𝜌𝑠 2 𝜌 2 𝑠 2 − 𝑠 2 𝑊 =

Taking 𝜌 = 3.142, the function to be maximized is 1 1 2500 2 3 × 3.142 × 𝑠 2 3.142 2 ×𝑠 2 − 𝑠 2 𝑊(𝑠) = 1 253.2 2 𝑊(𝑠) = 1.047𝑠 2 𝑠 2 − 𝑠 2

Let’s assume that we want to make a closed cone which will hold 10 𝑔𝑢 3 of water with minimum possible surface area to reduce the cost of the material. 1 Surface area, 𝑇 = 𝜌𝑠 2 + 𝜌𝑠 ℎ 2 + 𝑠 2 2 Volume V, = 10 𝑔𝑢 3 = 1 30 3 𝜌𝑠 2 ℎ ⟹ ℎ = 𝜌𝑠 2 Substituting the value of ℎ we get, 1 900 𝑇 = 𝜌𝑠 2 + 𝜌𝑠 2 𝜌 2 𝑠 4 + 𝑠 2 Taking 𝜌 = 3.142, we get the function, 1 𝑇(𝑠) = 3.142𝑠 2 + 3.142𝑠 91.17 2 𝑠 4 + 𝑠 2 We then find the minimum value of this function to get the minimum surface area that can hold 10 𝑔𝑢 3 .

Example A man wants to make a conical porous container to hold 2 𝑔𝑢 3 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 1 Surface area, 𝑇 = 𝜌𝑠 2 + 𝜌𝑠 ℎ 2 + 𝑠 2 2 1 6 Volume V, = 2 𝑔𝑢 3 = 3 𝜌𝑠 2 ℎ ⟹ ℎ = 𝜌𝑠 2 Substituting the value of ℎ we get, 1 𝑇 = 𝜌𝑠 2 + 𝜌𝑠 36 2 𝜌 2 𝑠 4 + 𝑠 2 Taking 𝜌 = 3.142, we get the function, 1 3.647 𝑇(𝑠) = 3.142𝑠 2 + 3.142𝑠 2 𝑠 4 + 𝑠 2

HOMEWORK A canvas has a surface area of 250 𝑔𝑢 3 . If this canvas is used to make a conical tent, write the volume of the tent as a function of its radius.

A NSWERS TO HOMEWORK 1 6331 2 𝑊(𝑠) = 1.047𝑠 2 𝑠 2 − 𝑠 2

THE END