JEE Advanced 2015 · Paper 1 · Mathematics Q47

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MathematicsInteger+4Question 47 of 60

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 $\dfrac{V}{250\pi}$ is

Your answer

Answer and worked solution

Official answer: 4

Material. With inner radius $r$ and inner height $h=V/(\pi r^2)$ ,
$\text{material}=\pi(r+2)^2(h+2)-V.$

Minimise. Differentiate in $r$ and set the result to zero:
$2(r+2)\left(\frac V{\pi r^2}+2\right)=(r+2)^2\frac{2V}{\pi r^3}\implies\frac V{\pi r^2}+2=\frac V{\pi r^2}+\frac{2V}{\pi r^3}.$
This gives $V=\pi r^3$ .

Value. With $r=10$ , $V=1000\pi$ , so $\dfrac V{250\pi}=4$ .

Answer: 4

Topics:Maxima and minima

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JEE Advanced 2015 · Paper 1, Mathematics question 47