a positive quantity, showing that, when x=2, y assumes a minimum value, whilst, when X d2y -2, 3' dx2 which is negative, showing that, for this value of x, y assumes a maximum value. where n and p are positive integers, show that x=a gives neither maximum nor minimum values of y, but that x=b gives a minimum. It will be clear from this example that neither maxima nor minima dy values can arise from the vanishing of such factors of as have even dx has a maximum value when x=4 and a 5. To show that a triangle of maximum area inscribed in any oval curve is such that the tangent at each angular point is parallel to the opposite side. If PQR be a maximum triangle inscribed in the oval, its vertex P lies between the vertices L, M of two equal triangles LQR, MQR inscribed in the oval. Now, the chord LM is parallel to QR and the tangent at P is the limiting position of the chord LM, which proves the proposition. It follows that, if the oval be an ellipse, the medians of the triangle are diameters of the curve, and therefore the centre of gravity of the triangle is at the centre of the ellipse. 6. Show that the sides of a triangle of minimum area circumscribing any oval curve are bisected at the points of contact; and hence that, if the oval be an ellipse, the centre of gravity of such a triangle coincides with the centre of the ellipse. 7. To find the path of a ray of light from a point A in one medium to a point B in another medium supposing the path to be such that the least possible time is occupied in passing from A to B, and that the velocity of propagation of light changes from v to v'on passing the boundary separating the media. [FERMAT'S PROBLEM.] We shall, for simplicity, consider A and B to lie in the plane of the paper, and the separating surface of the media to be cylindrical with its generators perpendicular to the plane of the paper. Let OPP' be the section of the separating surface by the plane of the paper, and let APB, AP'B be two contiguous paths from A to B. Then, if the times in these two paths be equal, the quickest path and therefore, if in the limit the incident ray AP and the refracted ray PB make angles i, i' respectively with the normal at P, we obtain sin i V sin i'v' thus proving Snell's well known law of refraction. 350. Analytical Investigation. We now proceed to investigate the conditions for the existence of maxima and minima values from a purely analytical point of view. It appears from the definition given of maxima and minima values that as x increases or decreases from the value a through any small but finite interval h, if p(x) be always less than p(a), then p(a) is a maximum value of p(x); and that if p(x) be always greater than p(a), then 4(a) is a minimum value of p(x). We shall assume in the present article that none of the derived functions we find it necessary to employ become infinite or discontinuous for the particular values discussed of the independent variable. We then have by Lagrange's modification of Taylor's Theorem h2 and p(x− h) — p(x) = −hp′(x)+2′′ (x — Oh) 2! (A) And when h is made sufficiently small the sign of the right-hand side of each equation, and therefore also of the left-hand side, is ultimately dependent upon that of hp'(x), that being the term of lowest degree in h. Hence and p(x+h)−p(x) have in general opposite signs. For a maximum or minimum value, however, it has been explained above that these expressions must, when h is taken small enough, have the same sign. It is therefore necessary that p'(x) should vanish, so that the lowest terms of the right-hand sides of the equations (A) should depend upon an even power of h. '(x)=0 is therefore an essential condition for the occurrence of a maximum or minimum value. Let the roots of this equation be a, b, c, Consider the root .... x= a. We may now replace equations (A) by the two equations h2 p(a+h)−p(a)='¿¡p′′(a)+ ̧ ̧p”(a+0 ̧h) h3 (B) h2 It is obvious now as before that the term "(a), being 2! that of lowest degree, governs the sign of the right and therefore also of the left side of each of equations (B); i.e., in general the signs of are the same as that of p′′(a). Hence if p′′(a) be negative p(a+h) and p(a−h) are both <4(a), and therefore 4(a) is a maximum value of p(x); while if "(a) be positive both p(a+h) and 4(a–h) are > 4(a), and therefore p(a) is a minimum value of p(x). But if it should happen that "(a) vanishes, equations (B) are replaced by 13 h4 p(a+h)-p(a)=8" (a) +48 (a+0h) h3 h4 p(a− h) — p(a) = - '(a)+'—8'"""(a — O2h) 4! and therefore when h is sufficiently small p(a+h)−p(a) l (C) are of opposite signs, and therefore there cannot be a maximum or minimum value of p(x) when x=ɑ unless ""(a) also vanish, in which case the sign of the right side of each equation depends upon that of p''''(a). And, as before, if this be negative we have a maximum value and if positive a minimum. Similarly, if several successive differential coefficients vanish when a is put equal to a, it appears that for a maximum or minimum value it is essential that the first not vanishing should be of an even order, and that if that differential coefficient be negative when xa a maximum value of p(x) is indicated, but if positive a minimum. Putting this =0 we obtain x=0, x=1, x=2. If x=1, "(x) is negative and therefore we have a maximum value; if x=2, 4′′(x) is positive and therefore this value of x gives a minimum value for p(x). If x=0, "(x) vanishes, so we must proceed further. Now ""(x)=60(12x2-18x+4), which does not vanish when x=0, so x=0 gives neither a maximum nor a minimum. 2. Show that x=0 gives a maximum value, and x=1 a minimum, x3 x2 for the function 3 2 3. Show that x=0 gives a maximum and x=1 a minimum for 4. Show that the expression sin30 cos attains a maximum value when 0=60°. 5. Illustrate geometrically the statement of Årt. 350 that in general p(x+h)− 4(x) and p(x− h) — 4(x) are of opposite sign. |