But from the Calculus we have the radius of curvature, Q= Further, the normal, MT, bisects the angle ATF; 2 A AM+MF 2√A2-B2 = AM r+r r MF' r = MF MF A2 N3 B P B=0. 2. To find by construction the corresponding point in the evolute. We have just found that r': P=9:r. Produce the normal TM, and also TF. On TM produced take Tp=P, and TL=r. Draw p F, and L X parallel to p F; then SOLUTION OF PROBLEMS IN MAXIMA AND MINIMA BY ALGEBRA. By RAMCHUNDRA, Late Teacher of Science, Delhi College. 1. To divide a given number into two such parts that their product may be the greatest possible. x, and a x Put the given number = a, one of the parts required consequently a-x= the other part. Therefore x (a — x) = 22= product = maximum=r. Therefore 22-ax=-r. Solving this quadratic equation we find x= Now it is evident that r cannot be greater than a2; for if it be so, the value of x becomes impossible; therefore the product or r, is greatest when a2=r; therefore x= † a 2' SECOND SOLUTION WITHOUT IMPOSSIBLE ROOTS. - 22, In the expression a x-x2, which is to become a maximum, let a x=y, where the value of y, determined by the condition of 2 a x-x2 being a maximum, will show whether it is positive, zero, or negative. We now find a x − x2=ay+22 — y2 — a y—4—¦2 — y2, α as 8/2 which is evidently a maximum when y = 0. Therefore x = 02 before. II. Of all right-angled plane triangles having the same hypothenuse, to find that whose area is the greatest possible. Let a = hypothenuse, x = base, y = perpendicular. Then, x2+ y2= a2, we shall have y = √a2x2, and consequently ху x 2 2 √a2x2= the area of the triangle = maximum, and consequently the square of the area, or (a2 x2 — x1) = max., and also four times this, or a2x2-x-max.= r. Therefore — a2x2-r. Solving this quadratic equation, we find and it is manifest that a2x2-24, or r cannot be greater than a*; therefore when r= maximum, we must have ra, therefore α x2=a2, and x = √2, and y = a y = √ a2 — x2 = √2* Hence it appears that the right-angled plane triangle contains the greatest area whose two sides, containing the right angle, are equal to each other. SECOND SOLUTION WITHOUT IMPOSSIBLE ROOTS. In the expression a2x2-x+, which is to become a maximum, let x2 = y2+ a2 2' which is evidently a maximum, when y = 0, and therefore 22: III. To bisect a triangle by the shortest line. = Let ABC be the given triangle, and DQ the shortest line required. Also, let CD=x, CQ=y, DQ=u, and a, b, c the three sides of the triangle, and the angle BCA. DM and BN are perpendiculars, drawn from the points D and B, on the line CA. Now, by similar triangles, we find ᎠᎷ BN = CD Св DM-x sin C, and BN CQX DM = sin C; therefore a sin C, and absin C 2 By Proposition 13, Book 2d, of EUCLID, we find Completing the square, and extracting the square root, we find, Now a2b2 is greater than a2 62 cos2 C; therefore in order that the value of 22 may not become impossible, we must have ab cos C + r ab; therefore ra ba b cos C, and therefore, when r= min., we must have IV. To find the least parabola which shall circumscribe a given circle. Since the parabola and the circle touch at E, therefore PE is a normal to the parabola, and PB is the subnormal = semi-parameter. Let PB =2; therefore the equation of the abola is y2=2zx. Also, 2z par Now the area of the parabola FAD ACX CF and CF √2z.AC. .. Area FADACX √2z. AC 2z.AC 2(r+z)3 Let r+z=y; therefore z=y-r. Therefore y T and therefore y3-uyur= 0. Let one of the negative roots —uy+ur= of this equation =-a, and therefore y+a must exactly divide the equation y―uyur=0. y+a) y3-uy+ur=0 (y2−ay + a2 — u = 0 . . . (B) |