A cylindrical container is to be made from certain solid material with the following constraints: It has a fixed inner volume of V mm3, has a 2 mm thick solid wall and is open at the top. The bottom of the container is a solid circular disc of thickness 2 mm and is of radius equal to the outer radius of the container. If the volume of the material used to make the container is minimum when the inner radius of the container is 10 mm, then the value of \frac{V}{250 \pi} is

Engineering

Mathematics

Maxima and Minima

Question

A cylindrical container is to be made from certain solid material with the following constraints: It has a fixed inner volume of V mm3, has a 2 mm thick solid wall and is open at the top. The bottom of the container is a solid circular disc of thickness 2 mm and is of radius equal to the outer radius of the container. If the volume of the material used to make the container is minimum when the inner radius of the container is 10 mm, then the value of $\frac{V}{250 π}$ is

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

$vm = {πr}_{E}^{2} h_{E} - {πr}_{I}^{2} h_{I} = π {(r + 2)}^{2} (h + 2) - V = π {(r + 2)}^{2} (\frac{V}{{πr}^{2}} + 2) - V$

$\frac{{dV}_{m}}{dr} = π \cdot 2 (r + 2) (\frac{V}{{πr}^{2}} + 2) + π {(r + 2)}^{2} (\frac{- 2 V}{{πr}^{3}}) = 0$

$\Rightarrow \frac{2 (V + 2 {πr}^{2})}{{πr}^{2}} = \frac{2 V (r + 2)}{{πr}^{3}}$

⇒ 2Vr + 4πr³ = 2Vr + 4V

⇒ V = πr³ = 1000π

$ ∴ \frac{V}{250 π} = 4 .$