Question

A container opened from the top and made up of a metal sheet is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of the milk which can completely fill the container at the rate of Rs. 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs. 8 per 100 cm². (Take π = 3.14)

Answer 1

Reqd. cost of milk = Vol. of the container × Rs. 20. 

The container is in the shape of the frustum of a cone whose, height (h) = 16 cm, radius of lower end (r₁) = 8 cm, radius of upper end (r₂) = 20 cm.

∴  Volume of the container = \(\frac{\pi}{3}\times h\times (r_1^2+r_2^2+r_1r_2)\)    [Using formula for frustum of a cone]

= \(\frac{3.14}{3} \times 16\times[(8)^2+(20)^2+160]\) 

= \(\frac{3.14}{3} \times 16\times[64+400+160]\) 

= \(\frac{3.14\times 16\times 624}{3}\) cm³  = \(\frac{3.14\times 16\times 624}{3\times 1000}\) L = 10.45 L (approx)

∴  Cost of milk =Rs. (10.45 × 20) = Rs. 209 (approx) 

Area of the metal sheet used to make the container

= Lateral surface area of the container + Area of the base = π l (r₁ + r₂) +π r₁² 

where, l = \(\sqrt{h^2+(r_2-r_1)}\)  ^{= \(\sqrt{16^2+(20-8)^2}\)  = \(\sqrt{256+144} =\sqrt{400}\) =20cm}

∴  Required area = 3.14 × 20 × 28 + 3.14 × 64 = 1758.4 + 200.96 = 1959.36 cm²

^∴ Cost of metal sheet = Rs. \(\big(\frac{8\times 1959.36}{100}\big)\) = Rs. 156.75 (approx).