# A container, open from the top and made up of a metal sheet is the form of frustum of a cone of height 30 cm with radii 30 cm and 10 cm. Find the cost of the milk which can completely fill container at the rate of Rs. 30 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs. 50 per 100 cm2. (π = 3.14)

Given.

Radii of frustum is 10 cm and 30 cm

Height of frustum is 30 cm

Rate of milk is Rs. 30 per litre

Rate metallic sheet is Rs . 50 per 100 cm2

Formula used/Theory.

CSA of frustum = π(r + R) × l

Volume of frustum = h[R2 + r2 + Rr]

Volume of Frustum = h[R2 + r2 + Rr]

= × 30cm × [(30 cm)2 + (10 cm)2 + 30 cm × 10 cm]

= 31.4 cm × [900cm2 + 100cm2 + 300cm2]

= 31.4 cm × 1300 cm2

= 40820 cm3

1 cm3 = litres

40820 cm3 = litres

= 40.820 litres

As milk is Rs. 30 per litre

For 40.82 litres

Rs. 30 × 40.82

= Rs. 1224.6

In frustum

L2 = height2 +

L2 = 302 + [30 – 10]2

L2 = 900 cm2 + 400 cm2 = 1300 cm2

L = √(1300 cm2) = 10√13 cm

CSA of Frustum = π(30 + 10)10√13 cm

= 3.14 × 40cm × 10√13cm

= 1256√13cm2

Area of base = πr2

= 3.14 × 10 × 10

= 314 cm2

Area of container = 1256√13 + 314

= 314[4√13 + 1]

= 314[4 × 3.6 + 1]

= 314 × 15.4

= 4835.6 cm2

Cost of 100 cm2 metal sheet = Rs. 50

Cost of 1 cm2 metal sheet = Rs. 0.5

Cost of 4835.6 cm2 metal sheet = Rs. 4835.6 × 0.5

= Rs. 2417.8

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