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# It is desired to construct a cylindrical vessel of capacity 500 cubic metres open at the top.What should be the dimensions of the vessel so that the material need is minimum, given that the thickness of the material used is 2 cm.r=(100π)1/3,h=5(100π)1/3r=(500π)1/3=hr=(1000π)1/3,h=(125π)1/3r=(20π)1/3,h=(100π)1/3

A
r=(100π)1/3,h=5(100π)1/3
B
r=(1000π)1/3,h=(125π)1/3
C
r=(500π)1/3=h
D
r=(20π)1/3,h=(100π)1/3
Solution
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#### Let r and h be the internal radius and height of the cylindrical vessel and let V denote the volume of the material. Now, V=πr2h⇒500=πr2h ⇒h=500πr2 ...(1)Now, S=(2πrh+πr2)2100 (As 2100 is thickness) =(1000r+πr2).150 For maximum or minimum , dsdr=(−1000r2+2πr)150=0 ⇒πr3=500∴πr3=500=V=πr2h or r=(500π)1/3=h Clearlyd2Sdr2=(4Vr3+2π).150=+ive∴ For minimum volume, r=(500π)1/3=h

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