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Question

An open tank is to be constructed with a square base and vertical sides so as to contain a given quantity of water. Show that the expenses of lining with lead with be least, if depth is made half of width.

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Solution

Let l, h, V and S be the length, height, volume and surface area of the tank to be constructed.Since volume, V is constant,l2h=Vh=Vl2 ...1Surface area, S=l2+4lhS=l2+4Vl From eq. 1dSdl=2l-4Vl2For S to be maximum or minimum, we must havedSdl=02l-4Vl2=02l3-4V=02l3=4Vl3=2VNow, d2Sdl2=2+8Vl3d2Sdl2=2+8V2V=6>0Here, surface area is minimum.h=Vl2Substituting the value of V=l32 in eq. 1, we geth=l32l2h=l2Hence proved.

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