47.If the dimensions are xft by xft by hft, then the totalamount of steel used is x214xhft2. Therefore,x214xh5108 and so h5}10842xx2}. The volume is givenby V(x)5x2h5}108x42x3}527x20.25x3. Then V9(x)52720.75x250.75(61x)(62x) and V0(x)521.5x. The critical point occurs at x56, and itcorresponds to the maximum volume because V0(x) ,0 forx .0. The corresponding height is }1084(26)623 ft. Thebase measures 6 ft by 6 ft, and the height is 3 ft.48.If the dimensions are xft by xft by hft, then we havex2h532 and so h5 }3x22}. Neglecting the quarter-inchthickness of the steel, the area of the steel used is A(x)5x214xh5x21 }12x8}. We can minimize the weightof the vat by minimizing this quantity. Now A9(x)52x2128x225 }x22}(x3243) and A0(x)521256x23. The critical point occurs at x54 andcorresponds to the minimum possible area because A0(x) .0 for x .0. The corresponding height is }3422} 52 ft. The base should measure 4 ft by 4 ft, and theheight should be 2 ft.
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