Ex. 1. To inscribe the greatest rectangle in a triangle. Ex. 2. Given the base and perpendicular of a triangle, to describe it so that the vertical angle may be a maximum. Ex. 3. To find the point D in the straight line CE, from which AB subtends the greatest angle. Ex. 4. x cos. (a - b) x sin. ¿ sin.20+ (a .. y = x2 sin.2 0 + (a (b x cos. 6) (b b) x sin. = 0 To find the least parabola which shall circumscribe a given eircle. Ex. 5. To divide an angle 24 into two parts, such that the mth power of the sine of one part by the nth power of the sine of the other part may be the greatest possible. m + n : m − n : . tan (4 + 0) + tan (p—0) : tan (ø + 0) — tan (4 — 0) .. sin 20 :: sin 24 m-n m + n : sin 20 sin 24; hence the two parts are known. Ex. 6. The four edges of a rectangular piece of lead, a inches in length, and b inches in breadth, are to be turned up perpendicularly, so as to form a vessel that shall hold the greatest quantity of water; how much of the edge must be turned up? and x = {a + b −√/a2 — ab +} gives the maximum vessel. EXAMPLES IN MAXIMA AND MINIMA. (1.) Of all triangles on the same base, having the same given perimeter, to find that whose surface is the greatest. (2.) Given the hypothenuse of a right-angled triangle, to determine the other sides, when the surface is the greatest possible. (3.) The whole surface of a cylinder being given = a2, to find its base and altitude, when the volume of the cylinder is a maximum. (4.) The volume of a cylinder = b3; find its base and altitude, when its whole surface is a minimum. (5.) Of all the squares inscribed in a given square, find that which is the least. (6.) Cut the greatest parabola from a given cone. (7.) Inscribe the greatest rectangle in a given ellipse. (8.) Find the longest straight pole that can be put up a chimney, when the height from the floor to the mantela, and the depth from front to back = b. (9.) AB is the diameter of a given semicircle; it is required to draw a chord PQ parallel to AB; so that if AQ and BP be joined intersecting in R, the triangle PQR may be a maximum. (10.) Inscribe the greatest cone in a sphere whose radius is a. ON THE METHOD OF LEAST SQUARES. In astronomical and physical researches, it is frequently required to determine the values of several quantities from a number of simple equations, and when the number of these equations is greater than the number of unknown quantities, they may be combined in a variety of ways, and each mode of combination will produce a different value of the unknown quantities. Hence it is a question of the highest importance to determine in what manner these equations are to be combined so as to give the values of the unknowns affected with the smallest probable errors, or in what way the values of these unknowns are to be found, so that each of the given equations may be satisfied with the greatest accuracy. Thus, for example, if from observation we have the four equations and it is required to find the values of x, y, z, we may pursue various methods and obtain various results for these unknowns. If the coefficient of a be made the greatest possible, while those of y and z are the smallest possible, we shall evidently have the most accurate value of x; because the value of x depends on those of y and z, and when their coefficients are the smallest possible, the terms in which y and z appear will then have the smallest influence on the value of x. In order, therefore, to obtain the most accurate value of x, we must change the signs of all the terms in eq. (4), and then by addition we get To make the coefficient of y a maximum, change the signs in eq. (1) and add, then we have Similarly for z, change the signs of eq. (2) and add, then we have Hence from (5), (6), (7), we have by the usual mode of elimination x = 2.4853, y= 3.5104, z = 1.9289. This method is practised in astronomy; but in point of accuracy it yields to the method of least squares, invented by Gauss, and to which modern astronomy owes much of its precision. Suppose then that e, e', e', . . . . are the errors of a series of observations, and that we have the equations ... ... &c. to determine the values of x, y, z, so that the errors e, e', e", e"", in reference to the whole of the observations shall be the least possible. Now if we were to take simply the sum of these errors, and put the differential coefficient of each of the variables equal to zero, we could not obtain an equation for the determination of the unknown quantity; but if we square each of the equations, we should have 12 e2+e12 +e112 +e!!112 + ...=x2 (a2 + a12 + ..)+2x {ah+a'h' +...+a(by+cz+...) +a'(by+cz+..)+a"(b"y+c"z+...) } + h2 + h12 + .... where the terms involving y and z are not written down, being exactly of the same form as the terms involving x. Let then, for the sake of brevity, u = e2 + e12 + e112 + e!!!2 + . .... Ax2+2Bx + C + .... and, therefore, putting the first differential coefficient of this equation equal to zero, we have, considering x alone as the variable, hence x (a2+a12+...)+ah+a'h'+...a(by+cz+...)+a'(b'y+c'z+...)+ &c.=0 or a (h+ax+by+cz)+a' (h'+a'x+by+cz)+ ..... =0 ... and therefore to form an equation that gives a minimum for any one of the unknown quantities, as x, we must multiply each equation of condition by the coefficient of the unknown quantity in that equation, taken with its proper sign, and equate the sum to zero. Proceed in the same manner for y, z, .. and we shall have as many equations of the first degree as there are unknown quantities, which may then be obtained by the usual mode of elimination. To apply this method to the preceding example, we have these equations:(1.) × (2.) X 3 (3.) × (4.) x 1 gives 1 3 + 30 y + 2z=0 15+ 9x+ by 1520 84 +16x+4y+ 16z= 0 14 + X- -3y and putting the sum of these equations - = 0, we obtain 3z0 (8.) Proceeding in a similar manner for y and z, we derive the equations 6x+15y + 2= 70 and from these three equations, (8), (9), (10), we have (9.) (10.) x = 2·4702, y = 3.5507, z = 1.9157. |