1. Day 168 – Cubical and cuboidal structures

2. Introduction
Cuboidal and cubical structures abound in real life. Packing boxes, cartons, and matchboxes are some examples of cuboidal structures. At times, these structures are designed to have the maximum volume possible or to have the minimum surface area. In this lesson, we will apply geometric methods to solve design problems of the cubical and cuboidal structures or object to satisfy physical constraints or minimize cost.

3. Vocabulary Domain of a function
This is the set of all independent variables for which a function will give real values of the output.

4. Consider a cuboid with the length 𝑙, width 𝑤 and height ℎ
Consider a cuboid with the length 𝑙, width 𝑤 and height ℎ. This cuboid has volume, 𝑉=𝑙𝑤ℎ The surface area of this cuboid is 𝑆.𝐴=2 𝑙𝑤 +2 𝑙ℎ +2 ℎ𝑤 . The volume of a cube with side 𝑙 is 𝑙 3 and its surface area is 6 𝑙 2 𝑙 𝑤 ℎ

5. We will consider a cuboid with a surface area of 50 𝑖𝑛 2 and a length of 5 𝑖𝑛. The volume of this cuboid is 𝑉=𝑙𝑤ℎ=5𝑤ℎ Its surface area is 𝑆.𝐴=50=2 𝑙𝑤 +2 𝑙ℎ +2 ℎ𝑤 . =2 5𝑤 +2 5ℎ +2 ℎ𝑤 50=10𝑤+10ℎ+2ℎ𝑤 Making ℎ the subject of the formula we get, ℎ= 25−5𝑤 5+𝑤 Substituting this in the volume equation we get, 𝑉(𝑤)=5𝑤( 25−5𝑤 5+𝑤 ) or 𝑉 𝑤 =125𝑤−25 𝑤 2 5+𝑤 −1

6. We then find the maximum point of the function in case we want to find the value of 𝑤 for the volume is maximum. At times we may want a cuboid with a specific volume and the maximum surface area possible. For instance, a cuboid with a volume of 100 𝑖𝑛 3 and a length of 5 𝑖𝑛. Let S be the surface area of the cuboid. 𝑆=2 𝑙𝑤 +2 𝑙ℎ +2 ℎ𝑤

7. Since 𝑙=5 𝑖𝑛, 𝑆=10𝑤+10ℎ+2 ℎ𝑤 Volume,𝑉=𝑙𝑤ℎ=5𝑤ℎ=100⟹ℎ= 20 𝑤 Substituting the value of h in the equation the surface area we have, 𝑆=10𝑤 ℎ +2 𝑤 ( 20 𝑤 ) 𝑆(𝑤)=10𝑤+ 200 𝑤 +40 To get the maximum surface area we find the maximum point of this function.

8. Example 1 A cuboid with a width 4 𝑖𝑛 has a surface area of 40 𝑖𝑛 2
Example 1 A cuboid with a width 4 𝑖𝑛 has a surface area of 40 𝑖𝑛 2 .Write the volume of the cuboid as a function of its length. Solution The volume of this cuboid is 𝑉=𝑙𝑤ℎ=4𝑙ℎ Its surface area is 𝑆.𝐴=40=2 𝑙𝑤 +2 𝑤ℎ +2 𝑙ℎ . =2 4𝑙 +2 4ℎ +2 𝑙ℎ 40=8𝑙+8ℎ+2𝑙ℎ Making ℎ the subject of the formula we get, ℎ= 20−4𝑙 4+𝑙

9. Substituting this in the volume equation we get, 𝑉(𝑤)=4𝑙( 20−4𝑙 4+𝑙 )

10. An open cuboid can be made by cutting out squares from its corners and folding up the sides. Consider a general case where a rectangle of length 𝑙 and width 𝑤 is made into a cuboid by cutting out squares of length 𝑥 and folding up the sides. The length of the cuboid is 𝑙−2𝑥, the width is 𝑤−2𝑥 and the height is 𝑥. 𝑙 𝑤 𝑥 𝑥
𝑙−2𝑥
𝑤−2𝑥

11. Its volume is 𝑥(𝑙−2𝑥)(𝑤−2𝑥). The surface area is S
Its volume is 𝑥(𝑙−2𝑥)(𝑤−2𝑥). The surface area is S.A= 𝑙−2𝑥 𝑤−2𝑥 +2𝑥 𝑙−2𝑥 +2𝑥(𝑤−2𝑥) Example 2 A rectangular piece of manila paper measures 14 𝑖𝑛 𝑏𝑦 8 𝑖𝑛. Squares of length 𝑥 𝑖𝑛 are cut out of the corners of this manila paper and sides folded up to make an open cuboid. Write the volume of the cuboid as a function of 𝑥 and determine its domain.

12. Solution The dimensions of the cuboid are 14−2𝑥 𝑖𝑛, 8−2𝑥 8−2𝑥 𝑖𝑛 and 𝑥 𝑖𝑛. Since the cuboid cannot have negative dimensions, 14−2𝑥>0, 8−2𝑥>0 and 𝑥>0 14−2𝑥>0 ⟹14>2𝑥⇒7>𝑥 8−2𝑥>0 ⟹8>2𝑥⇒4>𝑥 𝑥 lies between zero and 4. We can write 0<𝑥<4. This is called the domain of the function of the volume Volume 𝑉=𝑥 14−2𝑥 8−2𝑥 =4 𝑥 3 −44 𝑥 𝑥 Thus V(𝑥)=4 𝑥 3 −44 𝑥 𝑥

13. homework
A cuboid with a length of 4 𝑓𝑡 has a volume of 28 𝑓𝑡 3 . Write its surface area as a function of its width.

14. Answers to homework
𝑆 𝑤 =8𝑤+ 56 𝑤 +14

15. THE END