Given,
r1 = 20 cm,
r2 = 8 cm and
h = 16 cm
\[\therefore Volume\text{ }of\text{ }the\text{ }frustum\text{ }=\text{ }\left( \right)\times \pi \times h\left( r12+r22+r1r2 \right)\]
It is given that the pace of milk \[=\text{ }Rs.\text{ }20/liter\]
Thus, Cost of milk = 20×volume of the frustum
\[=\text{ }Rs.\text{ }209\]
Presently, incline tallness will be
Thus, CSA of the holder \[=\text{ }\pi \left( r1+r2 \right)\times l\]
Henceforth, the all out metal that would be needed to make holder will be \[=\text{ }1758.4\text{ }+\text{ }\left( Area\text{ }of\text{ }base\text{ }circle \right)\]
\[=\text{ }1758.4+201\text{ }=\text{ }1959.4\text{ }cm2\]
∴ Total expense of metal \[=\text{ }Rs.\text{ }\left( 8/100 \right)\text{ }\times \text{ }1959.4\text{ }=\text{ }Rs.\text{ }157\]