a milk container is made of metal sheet in the shape of frustum of cone whose volume is 104593/7cm^3 - Maths - Surface Areas and Volumes

Question

a milk container is made of metal sheet in the shape of frustum
of cone whose volume is 104593/7cm^3 the radii of its lower and
upper circular ends are 8cm and 20cm find the cost of the metal
sheet used in making the container at rate of Rs 1 40 per square
cm

Ajanta Trivedi · Accepted Answer

let the radii of lower end of the frustum be r = 8 cm.

let the radii of upper end of the frustum be R=20 cm

let the height of the frustum be h cm.

volume of the frustum = $\frac{π}{3} h (R^{2} + r^{2} + R r) = 10459 \frac{3}{7} = \frac{73216}{7}$

therefore substituting the value of R and r,

$\frac{22}{7} * \frac{1}{3} h (20^{2} + 8^{2} + 20 * 8) = \frac{73216}{7} h (400 + 64 + 160) = \frac{73216}{7} * \frac{7}{22} * 3 h * 624 = 9984 h = \frac{9984}{624} = 16 c m .$

total surface area of the container = $π (R + r) \sqrt{(R - r)^{2} + h^{2}} + π r^{2}$

$\frac{22}{7} (20 + 8) \sqrt{(20 - 8)^{2} + 16^{2}} + \frac{22}{7} * 8^{2} = \frac{22}{7} * 28 \sqrt{12^{2} + 16^{2}} + \frac{22}{7} * 64 = \frac{22}{7} * 28 \sqrt{144 + 256} + \frac{22}{7} * 64 = \frac{22}{7} (28 * \sqrt{400} + 64) = \frac{22}{7} (28 * 20 + 64) = \frac{22}{7} (560 + 64) = \frac{22}{7} * 624$

cost of 1 cm square metal sheet is 1.40 Rs.

cost of required sheet = $\frac{22}{7} * 624 * 1.40 = 2745.60 R s .$

a milk container is made of metal sheet in the shape of frustum of cone whose volume is 104593/7cm^3. the radii of its lower and upper circular ends are 8cm and 20cm. find the cost of the metal sheet used in making the container at rate of Rs.1.40 per square cm.