Determine the number of pieces of material required to make the tank. How much money is used to buy the required pieces of material? Should they use paint in jerry cans or tins? Justify your answer.

Kipenyo $D = 12 \text{ ft}$
Rediasi $r = \frac{D}{2} = \frac{12 \text{ ft}}{2} = 6 \text{ ft}$
Urefu $h = 2.5 \text{ m} \times 3.3 \text{ ft/m} = 8.25 \text{ ft}$

Eneo la duara moja (juu au chini):

Eneo la uso uliopinda:

Eneo la kipande kimoja cha nyenzo = $8 \text{ ft} \times 4 \text{ ft} = 32 \text{ ft}^2$

Idadi ya vipande kwa ajili ya juu:

Idadi ya vipande kwa ajili ya chini:

Idadi ya vipande kwa ajili ya uso uliopinda:

Jumla ya vipande = Vipande vya juu + Vipande vya chini + Vipande vya uso uliopinda

Idadi ya vipande vya nyenzo vinavyohitajika ni $\boxed{\text{18 vipande}}$.

Gharama kwa kipande kimoja = shs. 80,000
Idadi ya vipande = 18 (kutoka sehemu a)

Kiasi cha pesa kinachotumika kununua nyenzo ni shs. $\boxed{\text{1,440,000}}$.

Eneo la kupakwa rangi ni sehemu ya nje ya tanki ukiondoa sehemu ya chini. Hivyo ni eneo la juu na eneo la uso uliopinda.

• Jerry can: Eneo linalopakwa = $300 \text{ ft}^2$ Idadi ya jerry cans = $\left\lceil \frac{424.115 \text{ ft}^2}{300 \text{ ft}^2} \right\rceil = \lceil 1.4137 \rceil = 2 \text{ jerry cans}$
• Sealed tin: Eneo linalopakwa = $200 \text{ ft}^2$ Idadi ya sealed tins = $\left\lceil \frac{424.115 \text{ ft}^2}{200 \text{ ft}^2} \right\rceil = \lceil 2.1205 \rceil = 3 \text{ sealed tins}$

• Jerry cans: Gharama kwa jerry can = shs. 23,000 Gharama ya jumla ya jerry cans = $2 \times 23,000 \text{ shs.} = 46,000 \text{ shs.}$
• Sealed tins: Gharama kwa sealed tin = shs. 14,000 Gharama ya jumla ya sealed tins = $3 \times 14,000 \text{ shs.} = 42,000 \text{ shs.}$

Gharama ya jerry cans ni shs. 46,000.
Gharama ya sealed tins ni shs. 42,000.

Wanapaswa kutumia sealed tins.
Uthibitisho: Jumla ya gharama ya kutumia sealed tins (shs. 42,000) ni ndogo kuliko kutumia jerry cans (shs. 46,000) kwa eneo lote la kupakwa rangi.