Question

A container made of a metal sheet open at the top is of the form of frustum of cone, whose height is 16 cm and the radii of its lower and upper circular edges are 8 cm and 20 cm respectively. Find

(i) the cost of metal sheet used to make the container if it costs Rs 10 per 100 $c{m}^2$

(ii) the cost of milk at the rate of Rs 35 per litre which can fill it completely.

Answer 1

(i)

Radius ($r_1$) of upper end of container = 20 cm

Radius ($r_2$) of lower end of container = 8 cm

Height (h) of container = 16 cm

Slant height (l) of frustum, $l=\sqrt{(r_1-r_2{)}^2+h^2}=\sqrt{(20-8{)}^2+{16}^2}=\sqrt{{12}^2+{16}^2}=\sqrt{144+256}=\sqrt{400}=20cm$

Capacity of container = Volume of frustum

$=\frac{1}{3}\pi h(r_1^2+r_2^2+r_1{r}_2)=\frac{1}{3}\times 3.14\times 16\times ({20}^2+8^2+20\times 8)=\frac{1}{3}\times 3.14\times 16\times (400+64+160)=\frac{1}{3}\times 3.14\times 16\times 624=10449.92c{m}^3=10.45litres$

(ii) Cost of 1 litre milk = Rs 35

Cost of 10.45 litre milk = 10.45 $\times$ 35

= Rs 365.75

(i) Area of metal sheet used to make the container
$=\pi (r_1+r_2)l+\pi r_2^2=\pi (20+8)l+\pi 8^2=560\pi +64\pi =624\pi c{m}^2$

Cost of 100 $c{m}^2$ metal sheet = Rs 10

cost of $624\pi c{m}^2$ metal sheet =$\frac{624\times 3.14\times 10}{100}=195.94$

Therefore, the cost of the milk which can completely fill the container is

Rs 365.75 and the cost of metal sheet used to make the container is Rs 195.94.