IMPLICIT FUNCTIONS. 351. In the case in which the quantity y, whose maximum and minimum values are the subject of investigation, appears as an implicit function of x, and cannot readily be expressed explicitly, we may proceed as follows: Let the connecting relation between x and y be therefore include the required maxima and minima. Differentiating equation (2) we have (3) ǝ2p dy dy ap d2y =0,... (4) 2 (away + du) + Ə2, Ə2 dy + and, remembering that dx Substituting the values of x and y derived from equa d2y tions (3) we can test the sign of and thus dis criminate between the maxima and minima values. The case in which this test fails, viz., when 2 =0 for the values of x and y deduced by equations (3), is complicated owing to the complex nature of the general d3y d+y formulae for and dx3 dx4 Ex. Find the maximum and minimum ordinates of the curve i.e., Combining this with the equation to the curve we obtain which presents the additional solution y=a3/4, x=a3/2. (1) Hence the points at which maxima or minima ordinates may exist have for their co-ordinates (0, 0) and (a3/2, a 3/4). and is negative, and therefore at this point y has a maximum value. At the point x=0, y=0, the formulae for dy and dy, both dx dx2, become indeterminate, and we have to investigate their true values. showing that the ordinate y has for this point a minimum value. SEVERAL DEPENDENT VARIABLES. 352. Suppose the quantity u, whose maxima and minima values are the subject of investigation, to be a function of n variables x, y, z, etc., but that by virtue of n-1 relations between them there is but one variable independent, say x. We may now, from the n-1 equations, theoretically find the n-1 dependent variables y, z, ... in terms of x, and suppose that by substitution u is expressed as a function of the one independent variable X. The methods of the preceding articles can now be applied. It is often, however, inconvenient, even if possible, actually to eliminate the n-1 dependent variables y, z, etc., and it is not necessary that this should be immediately done. Suppose, for instance, u=4(x, y, z) a function such as the one discussed, a the independent variable, y and z dependent variables connected with x and y by the relations an equation in x, y, z which, with_u=4(x, y, z), F1=0, and F2=0, will serve to find x, y, z and u. 2 Again, by differentiating equations (1), (2), (3), and dy dz d2y d2z eliminating dx' dx' dx2 dx2 we may deduce the value of d2u dx2 and test its sign for the values of x, y, z found. Ex. A Norman window consists of a rectangle surmounted by a semicircle. Given the perimeter, show that, when the quantity of light admitted is a maximum, the radius of the semicircle must equal the height of the rectangle. [TODHUNTER'S DIFF. CALC., p. 214, Ex. 30.] Let y be the height and 2x the breadth of the rectangle, then the area of the window is given by Choose x to be the independent variable. Then we have, since and therefore the radius of the semicircle is equal to the height of the rectangle. To test whether this result gives a maximum value to A we have =π+4dy + 2x dxv d2P d2y =0=2 dx2 dx2 d2 A dx2 and is therefore negative. Hence the relation found, viz., x=y, indicates a maximum value of the area. 353. In the solution of such questions as the foregoing it is frequently unnecessary to employ any test for the discrimination between the maxima and minima, since it is often sufficiently obvious from geometrical considerations which results give the maxima values and which give the minima. 354. Function of a Function. Suppose z=f(x), where x is capable of assuming all possible values, and let y=F(2); then it appears that dy_dy dz since F''(z)ƒ′(x), |