Class 09 PROJECT 02 SURFACE AREAS AND VOLUMES OF CUBOIDS

 PROJECT 02

SURFACE AREAS AND VOLUMES OF CUBOIDS

BACKGROUND 

Cuboidal objects are quite useful in our daily life and often we need to know their surface areas and volumes for different purposes. Sometimes, it appears that if there is an increase in the surface area of a cuboid, then its volume will also increase and vice-versa. The present project is a step towards knowing the truth about this statement. 

OBJECTIVE 

To explore the changes in behaviours of surface areas and volumes of cuboids with respect to each other. 

DESCRIPTION 

(A) Cuboids with equal volumes Let us consider some cuboids with equal volumes, having the following dimensions: 

(i) l = 12 cm, b = 6 cm and h = 3 cm 

(ii) l = 6 cm, b = 6 cm and h = 6 cm 

(iii) l = 9 cm, b = 6 cm and h = 4 cm

 (iv) l = 8 cm, b = 6 cm and h = 4.5 cm.

 Now, we calculate the surface area of each of the above cuboids, using the formula 

surface area = 2 (lb + bh + hl). 

For (i), surface area = 2 (12 × 6 + 6 × 3 + 3 × 12) cm2 = 252 cm2 

For (ii), surface area = 2 (6 × 6 + 6 × 6 + 6 × 6) cm2 = 216 cm2 → Minimum 

For (iii), surface area = 2 (9 × 6 + 6 × 4 + 4 × 9) cm2 = 228 cm2 

For (iv), surface area = 2 (8 × 6 + 6 × 4.5 + 4.5 × 8) cm2 = 222 cm2

We note that volume of each of the above cuboids = 12 × 6 × 3 cm3 = 6 × 6 × 6 cm3 = 9 × 6 × 4 cm3 = 8 × 6 × 4.5 cm3 = 216 cm3 . 

We also note that surface area of the cuboid is minimum, 

in case (ii) above, when the cuboid is a cube. 

(B) Cuboids with equal surface areas Let us now consider some cuboids with equal surface areas, having the following dimensions: 

(v) l = 14 cm, b = 6 cm and h = 5.4 cm 

(vi) l = 10 cm, b = 10 cm and h = 4.6 cm 

(vii) l = 8 cm, b = 8 cm and h = 8 cm

 (viii) l = 16 cm, b = 6.4 cm and h = 4 cm 

Now, we calculate the volume of each of the above cuboids

, using the formula volume = l × b × h 

For (v), volume = 14 × 6 × 5.4 cm3 = 453.6 cm3 

For (vi), volume = 10 × 10 × 4.6 cm3 = 460 cm3 

For (vii), volume = 8 × 8 × 8 cm3 = 512 cm3 → Maximum 

For (viii), volume = 16 × 6.4 × 4 cm3 = 409.6 cm3 

We note that surface area of each of the above

 cuboids = 2 (14 × 6 + 6 × 5.4 + 5.4 × 14) cm2 

= 2 (10 × 10 + 10 × 4.6 + 4.6 × 10) cm2

 = 2 (8 × 8 +8 × 8 + 8 × 8) cm2 

= 2 (16 × 6.4 + 6.4 × 4 + 4 × 16) cm2 = 384 cm2. 

We also note that volume of the cuboid is maximum, 

in case of (vii), when the cuboid is a cube

CONCLUSION 

The statement that if there is an increase in the surface area of cuboid, then its volume also increases and vice versa is not true. 

In fact, we have:

 (i) Of all the cuboids with equal volumes, the cube has the minimum surface area. 

(ii) Of all the cuboids with equal surface areas, the cube has the maximum volume. 

APPLICATION 

Project is useful in preparing packages with maximum capacity at minimum cost.