. Applied calculus; principles and applications . Take / (x) = x^ (a^ - x^) = a^x^ - x^ [by Art. 87]; / (x) =2aH-4:X^ = 2x (a^ -2x^) = 0; .-. x = 0, a/V2;J {x) =2a^-12x^; f (0) = 2 a^; /. / (0) = 0 is ) = 2o? - 6a2 = -4a2; .-. /(a/V2) = A = 4 V^ = 2 a2 is the area of the maximum rectangle, which is a — By Geometry without the Calculus method: A =2ay = 2aVa^ - x-]j^o = 2a2, since the radical quantity is evidently greatest when x = 2. The strength of a beam of rectangular cross section varies as thebreadth 6 and as d^, the square of the depth. Find the dimension
. Applied calculus; principles and applications . Take / (x) = x^ (a^ - x^) = a^x^ - x^ [by Art. 87]; / (x) =2aH-4:X^ = 2x (a^ -2x^) = 0; .-. x = 0, a/V2;J {x) =2a^-12x^; f (0) = 2 a^; /. / (0) = 0 is ) = 2o? - 6a2 = -4a2; .-. /(a/V2) = A = 4 V^ = 2 a2 is the area of the maximum rectangle, which is a — By Geometry without the Calculus method: A =2ay = 2aVa^ - x-]j^o = 2a2, since the radical quantity is evidently greatest when x = 2. The strength of a beam of rectangular cross section varies as thebreadth 6 and as d^, the square of the depth. Find the dimensions ofthe section of the strongest beam that can be cut from a cylindrical 128 DIFFERENTIAL CALCULUS log whose diameter is 2 a. Strength oc bd^; .. strength = kbd^, wherefc is a constant; let fih)=b (4a2 - 62) = 4,a% - b^; V-3(2a).2a „, 16a3 /(6) =4a2-362=0; 6 = -^, d V3 / (6) = ^ (4 a2 - ^) = -r^ . = (4 a^) = Hence, the rectangle may be laid off on the end of the log by drawinga diameter and dividing it into three equal parts; from the points ofdivision drawing perpendiculars in opposite directions to the circum-ference and joining the points of intersection with the ends of thediameter, as in the figure. The strength of the beam is about ofthat of the log, but it is the strongest beam of rectangular section. 3. The stiffness of a rectangular beam varies as the breadth 6 and asd^, the cube of the depth. Find the dimensions of the stiff est beam
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