The problem seeks the dimension of the cylindrical tin can that is capable of holding 14 cubic inches of liquid, and has the smallest surface area. Before we can determine the dimension, let us consider the volume, V, of this cylindrical tin can, which is the product of the area of the circular base with radius r, and the height h of the cylinder, and the surface area of the tin can, A, which is the area of the circular base, the circular top, and around the sides. V= πr2h (1) 14= πr2h h= 14πr2 (2) A=2πr2+ ...
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