Question

A milk container is made of metal sheet in the shape of frustum of a cone whose volume is $10459\frac{3}{7}c{m}^3.$ The radii of its lower and upper circular ends are 8 cm and 20 cm respectively. Find the cost of metal sheet used in making the container at the rate of Rs 1.40 per $c{m}^2$.

Answer 1

We have,

Radius of the upper end, R = 20 cm and

Radius of the lower end, r = 8 cm

Let the height of the container be h.

As,

Volume of the container = $10459\frac{3}{7}c{m}^3.$
= $\frac{73216}{7}c{m}^3$

⇒ $\frac{1}{3}\pi h(R^2+r^2+Rr)$= $\frac{73216}{7}$

⇒ $\frac{1}{3}\times \frac{22}{7}\times h\times ({20}^2+8^2+20\times 8)$ =$\frac{73216}{7}$

⇒ $\frac{22h}{21}\times (400+64+160)$ = $\frac{73216}{7}$

⇒ $\frac{22h}{21}\times 624=\frac{73216}{7}$

⇒ h = $\frac{(73216\times 21)}{(7\times 22\times 624)}$

⇒ h = 16 cm

Also,

The slant height of the container,
l = $\sqrt{(R-r{)}^2+h^2}$

= $\sqrt{(20-8{)}^2+{16}^2}$

= $\sqrt{{12}^2+{16}^2}$

= $\sqrt{144+256}$

= $\sqrt{400}$

= 20 cm

Now,

Total surface area of the container = πl(R+r)+ $\pi r^2$+ $\pi R^2$

= $\frac{22}{7}\times 20\times (20+8)+\frac{22}{7}\times 8\times 8$+ $\frac{22}{7}\times 20\times 20$

= $\frac{22}{7}\times 20\times 28+\frac{22}{7}\times 64$+ $\frac{22}{7}\times 400$

= $\frac{22}{7}\times (560+64+400)$

= $\frac{22}{7}\times 1024$

=$\frac{22528}{7}c{m}^2$

So, the cost of metal sheet =$1.4\times \frac{22528}{7}$=2745.60

Hence, the cost of the metal sheet used for making the milk container is Rs. 4505.60.