Taylor Series and stationary points

Chapter 5 Taylor Series and stationary points 5.1 Taylor Series The surface z = f(x, y) and its derivatives can give a series approximation for f(x, y) about some point (x 0, y 0 ) as illustrated in Figure 5.1 Let x = x 0 +ht and y = y 0 + kt, where h and k are constants. This then gives the parametric form of the straight line joining (x, y) and (x 0, y 0 ). Note that we can express x = x(t) and y = y(t). Therefore, z = f(x(t), y(t)) = z(t). So they are all functions of the parameter t. We can expand z(t) about t = 0, which corresponds to the point (x 0, y 0 ), in the usual Taylor series as z(t) = z(0) + ( ) dz + t2 dt t=0 2! ( d 2 ) z dt 2 + t=0 This is exactly the expression we used earlier in the course. However, the problem lies in calculating the derivatives because we know z as a function of x and y and not explicitly as a function t. At t = 0, x = x 0 and y = y 0 and so z(0) = f(x 0, y 0 ). So we are expanding about the point (x 0, y 0 ) parametrically. To evaluate (dz/dt) we use the chain rule. For a general function φ(x(t), y(t)), we have dφ dt = ( ) φ dx x dt + ( ) φ dy dt, (5.1) where Thus, ( ) dz t = dt t=0 dx dt = h, dy dt = k. ( ) th + x (x 0,y 0) ( ) tk. (x 0,y 0) 65

66 CHAPTER 5. TAYLOR SERIES AND STATIONARY POINTS Figure 5.1: The general point (x, y) and the point (x 0, y 0 ) about which the Taylor series is formed. Eliminating ht and kt in terms of x x 0 and y y 0, this can be expressed as ( ) ( ) ( ) dz t = (x x 0 ) + (y y 0 ). dt t=0 x (x 0,y 0) (x 0,y 0) Now we need to consider the second derivatives in the next term of the Taylor series. d 2 z dt 2 = d ( ) dz = d ( ) dt dt dt x h + d ( ) k. dt To evaluate the right hand side, we replace / x h by φ and use the previous result. Then we evaluate the second term by replacing / k by φ and again use (5.1). Thus, we get d 2 z dt 2 = { ( x x { + x ( ) h dx dt + ( x ) k dx dt + ) h dy dt ( } ) k dy dt = 2 f x 2 h2 + 2 f x hk + 2 f x hk + 2 f 2 k2 }

5.1. TAYLOR SERIES 67 This gives t 2 2! ( d 2 z dt 2 ) = h 2 2 f x 2 + 2hk 2 f x + k2 2 f 2. t=0 = 1 2! [ ] t 2 h 2 2 f x 2 + 2(ht)(kt) 2 f x + t2 k 2 2 f 2. (x 0,y 0 ) This can be written in a shorthand notation as t 2 ( d 2 ) [ z 2! dt 2 = t2 h 2! x + k ] 2 f (x0,y0). t=0 The square bracket terms can be expanded by the binomial expansion first before operating on the function f to calculate the derivatives. Replacing ht = (x x 0 ) and kt = (y y 0 ) gives our final expression for the Taylor series for a function in two variables, as f(x, y) = f(x 0, y 0 ) + + 1 2! [ (x x 0 ) x + (y y 0) ] [ (x x 0 ) 2 2 f x 2 + 2(x x 0)(y y 0 ) 2 f x + (y y 0) 2 2 f 2 ] + (5.2) All the derivatives are evaluated at (x 0, y 0 ). Using our shorthand notation we can express this as [ f(x, y) = f(x 0, y 0 ) + (x x 0 ) ] f + [ 1 (x x 0 ) 2! x + (y y 0) x + (y y 0) ] 2 f + Remember to expand the square term by the binomial expansion before working out the derivatives. Although we will very rarely continue beyond the second derivative terms, note that the shorthand notation allows us express the higher derivatives in a compact form. Thus, the third derivative gives t 3 d 3 z 3! dt 3 = 1 [ (x x 0 ) 3! x + (y y 0) ] 3 f. Again we must expand the square brackets by the binomial expansion before evaluating the derivatives. Example 5.1 Calculate the Taylor series about (0, 0) for f(x, y) = sin(x)e y.

68 CHAPTER 5. TAYLOR SERIES AND STATIONARY POINTS Evaluated at (0, 0) f(x, y) = sin(x)e y 0 x = cos(x)ey 1 = sin(x)ey 0 2 f x = sin(x)e y 2 0 2 f = sin(x)e y 2 0 2 f x = cos(x)ey 1 Table 5.1: All the first and second order derivatives evaluated at (0, 0). This could be done by expanding each factor as a function of one variable and then multiplying the answers together. However, it is our aim to illustrate the method for a function of two variables. Thus, remembering that the derivatives are evaluated at (0, 0), we have f(x, y) = f(0, 0) + x x + y + 1 ( ) x 2 2 f 2! x 2 + 2xy 2 f 2 + y2 x 2 + So we need to evaluate all the first and second derivatives. This is done first before substituting the numbers into the equation above. Thus, sin(x)e y 0 + [1 x + 0 y] + 1 2! [ x2 0 + 2xy 1 + y 2 0 ]. Simplifying we get sin(x)e y x + xy = x(1 + y) 5.2 Stationary Points When considering function of one variable, an important application of calculus is to determine the maximum and minimum values. This is done by locating the stationary points where the derivative is zero and then determine the nature of the stationary points by looking at the sign of the second derivatives. The same approach is used in studying functions of two variables. So let us revise the one variable case in detail before progressing to more independent variables. If y = f(x), then a necessary condition for a maximum or minimum (or more generally stationary points) is dy dx = f (x) = 0. For a stationary point at x 0, the Taylor series gives y = f(x) = f(x 0 ) + (x x 0 )f (x 0 ) + 1 2 (x x 0) 2 f (x 0 ) +

5.2. STATIONARY POINTS 69 Since, f (x 0 ) = 0 at a stationary point, the second term (linear in (x x 0 )) vanishes. Assuming that f (x 0 ) 0 and (x x 0 ) sufficiently small, f(x) f(x 0 ) 1 2 (x x 0) 2 f (x 0 ). If f (x 0 ) is positive, then f(x) > f(x 0 ) in the vicinity of x 0 and so f has a minimum at x 0. Conversely, if f (x 0 ) is negative, then f(x) < f(x 0 ) and f has a local maximum. In general, the nature of the stationary (or turning) points depends upon the first non-zero derivative. Thus, if f (x 0 ) = 0 and f (x 0 ) 0, then turning point is a maximum if f (x 0 ) < 0 and a minimum if f (x 0 ) > 0. If f (x 0 ) = f (x 0 ) = 0, then f(x) f(x 0 ) 1 3! (x x 0) 3 f (x 0 ). Regardless of the sign of the third derivative, f(x) f(x 0 ) has a different sign for x either side of x 0. Thus, f(x) has a point of inflexion at x 0. We can extend this to the case where the first three derivatives are zero and the fourth derivative is non-zero. The same approach will determine the local nature of the turning point. Similar concepts apply to functions of two variables, f(x, y). f(x, y) has a maximum at (x 0, y 0 ) if f(x, y) < f(x 0, y 0 ) in the vicinity of (x 0, y 0 ). Similarly, if f(x, y) > f(x 0, y 0 ), then f has a minimum at (x 0, y 0 ). Let there be a turning point at (x 0, y 0 ) as illustrated in Figure 5.2. Along the line y = y 0, f(x, y 0 ) is just a function of one variable and the condition for a turning point is simply ( ) = 0. x (x 0,y 0 ) Similarly, along the line x = x 0, f(x 0, y) must satisfy ( ) = 0. (x 0,y 0) Any point (x 0, y 0 ) where / x = / = 0 is called a stationary or critical point of the function f(x, y). Evidently, the tangent plane, T, is horizontal at (x 0, y 0, z = f(x 0, y 0 )) and f has no horizontal components. Example 5.2 Let z = f(x, y) = x 2 + y 2. x = 2x, = 2y. Therefore, at (0, 0) both derivatives are zero and the origin is a stationary point. By inspection, (0, 0) is a minimum.

70 CHAPTER 5. TAYLOR SERIES AND STATIONARY POINTS Figure 5.2: There is a maximum turning point at (1, 1). Example 5.3 Let z = f(x, y) = x 2 y 2. x = 2x, = 2y. Again the origin is a stationary point. But is it a maximum or a minimum? Obviously, along y = 0, z = x 2 and so the origin is a minimum. Bur, along x = 0, z = y 2 and the origin is a maximum. Note that z = 0 everywhere along the two lines y = x and y = x. These lines separate regions where z < 0 from regions where z > 0. This is shown in Figure 5.3. Thus, if x 2 > y 2, then z > 0 and if x 2 < y 2, z < 0. Therefore, the sign of [f(x, y) f(x 0, y 0 )] changes with (x, y) in the vicinity of (x 0, y 0 ). This is neither a maximum nor a minimum but a saddle point.

5.3. TESTING THE NATURE OF A STATIONARY POINT 71 Figure 5.3: Regions where z is positive are indicated by the shading. 5.3 Testing the nature of a stationary point Recall the Taylor series for f(x, y), [ f(x, y) = f(x 0, y 0 ) + t h x + k ] [ ] + t2 h 2 2 f 2! x 2 + 2hk 2 f x + k2 2 f 2 +, (5.3) where ht = x x 0 and kt = y y 0. If (x 0, y 0 ) is a stationary point / x = / = 0 and so (5.4) becomes where the notation f(x, y) f(x 0, y 0 ) = t2 [ h 2 f xx + 2hkf xy + k 2 ] f yy +, (5.4) 2! f xx = 2 f x 2, f xy = 2 f x, f yy = 2 f 2, is used. The condition for a minimum of f is that f(x, y).f(x 0, y 0 ) for all (x, y) near (x 0, y 0 ). This means that the coefficient of t 2 must be positive for all (x, y) or equivalently for all values of h and k. Similarly, the condition for a maximum

72 CHAPTER 5. TAYLOR SERIES AND STATIONARY POINTS is that the coefficient is negative. If the coefficient is zero, the expansion (5.3) does not help and we need to consider the third derivatives. Now consider the coefficient of t 2 is more detail. C = h 2 f xx + 2hkf xy = k 2 f yy. (5.5) Several cases exist and we will assume that f x = f y = 0 at (x 0, y 0 ) automatically. 5.3.1 Case 1 f xx = f yy = 0, but f xy 0 The coefficient reduces to 2hkf xy and will change sign as h and k change sign. Thus, f has neither a maximum nor a minimum but a saddle point. 5.3.2 Case 2 Suppose f xx 0 at (x 0, y 0 ). We will complete the square in h to obtain a positive term and an extra term. It is the sign of the extra term that determines the nature of the stationary point. Firstly, since f xx 0 we express C as Hence, we have C = = C = ( h 2 f xx + 2hkf xy + k 2 f yy ) f xx f xx. 1 [ h 2 fxx 2 + 2hkf xy f xx + k 2 f yy + { k 2 fxy 2 k 2 f 2 }] xy f xx 1 [ (hf xx + kf xy ) 2 + k 2 ( f xx f yy f f xy) ] 2. xx So we have completed the square and obtained a term (hf xx + kf xy ) 2, that is always positive but we can make it zero by choosing suitable values for h and k. Remember that simply defines the path. The other term k 2 ( f xx f yy f 2 xy), is the crucial term for determining the nature of the stationary point. Define = f xx f yy f 2 xy, at (x 0, y 0 ). (5.6) The sign of determines the nature of the stationary point in a similar way to the use of the second derivative for a function of one variable. If > 0, the sign of C is that of f xx. Hence, we have If > 0, and f xx > 0, C > 0, (x 0, y 0 ) is a minimum. If > 0, and f xx < 0, C < 0, (x 0, y 0 ) is a maximum.

5.3. TESTING THE NATURE OF A STATIONARY POINT 73 If < 0, then C = 1 [ ] (hf xx + kf xy ) 2 k 2. f xx If k = 0, then the square bracket is positive. However, if k 0, but chosen so that hf xx + kf xy = 0, then the square bracket is negative. Thus, C changes sign with the choices of h and k. So we have a saddle point. 5.3.3 Case 3 Suppose f yy 0 at (x 0, y 0 ). We can rewrite C as 1 [ ] (hf xy + kf yy ) 2 + h 2. f yy This can be verified by expanding to show that it is equivalent to the original expression for C. Hence, we have the same conclusions based upon the sign of as in Case 2. 5.3.4 Summary of Testing Compute = ( 2 ) f x 2 ( 2 ) ( f 2 ) 2 f 2, at (x 0, y 0 ). x 1. = 0 means that the test is inconclusive and we need a higher order Taylor series. 2. < 0 means that the stationary point is a saddle point. 3. > 0. In this case it is clear that can only be positive if the product 2 f/ x 2 2 f/ 2 is positive. Therefore, either both 2 f/ x 2 and f / 2 are positive or they are both negative. Example 5.4 > 0 and 2 f/ x 2 < 0 means that the stationary point is a maximum. > 0 and 2 f/ x 2 > 0 means that the stationary point is a minimum. Find the location and nature of the stationary points of z = f(x, y) = x 3 6xy + y 3. First the location is given by setting both first derivatives to zero. x = 3x2 6y = 0, y = 1 2 x2, and ( ) 2 1 = 6x + 3y2 = 0, 6x + 3 2 x2 = 0.

74 CHAPTER 5. TAYLOR SERIES AND STATIONARY POINTS Solving gives 2x + 1 4 x4 = 0, x = 0 or x = 2. So there are two stationary points at x = 0, y = 0 and x = 2, y = 2. determine their nature we calculate all the second derivatives. Therefore, we have f x 2 = 6x, 2 f 2 = 6y, 2 f x = 6. To = f xx f yy (f xy ) 2 = 36xy 36. At (0, 0), = 36 < 0 and so this is a saddle point. At (2, 2), = 36 2 2 36 = 108 > 0 and f xx = 12 > 0 (note that f yy is also positive) and so this is a minimum. Example 5.5 Determine the location and nature of the stationary points of The first derivatives give f(x, y) = x 4 + y 4 2(x y) 2. x = 4x3 4(x y), = 4y3 + 4(x y). Hence, the stationary points satisfy the pair of equations Add the two equations together to get 4x 3 4x + 4y = 0, 4y 3 + 4x 4y = 0. x 3 + y 3 = 0, x = y. Now substitute back into one of the equations to get 4x 3 8x = 0, x 3 2x = 0, x(x 2 2) = 0. This implies there are three stationary points at (0, 0), (+ 2, 2) and ( 2, + 2). Now to test their nature we need the second derivatives. 2 f x 2 = 12x2 4, 2 f 2 = 12y2 4, 2 f x = 4. Thus, we calculate = f xx f yy (f xy ) 2 = (12x 2 4) (12y 2 4) ( 4) 2 at each of the stationary points.

5.4. MAXIMA AND MINIMA SUBJECT TO A CONSTRAINT: LAGRANGE MULTIPLIERS75 At (+ 2, 2), = (24 4) (24 4) 16 > 0 and f xx = 20 > 0 and so this is a minimum. At ( (2), + 2), = (24 4) (24 4) 16 > 0 and f xx = 20 > 0 and so this is again a minimum. At the origin (0, 0), = ( 4) ( 4) 16 = 0. In this case the test is inconclusive and further analysis is needed. If we assume that x and y are small, then terms like x 4 and y 4 are small and can be neglected when compared to squared terms like 2(x y) 2. Thus, provided that this term does not vanish, namely when x = y, we have f(x, y) f(0, 0) = 2(x y) 2 < 0. However, if we choose a path such that x = y, then f(x, y) f(0, 0) = x 4 + x 4 > 0. Since f(x, y) f(0, 0) changes sign depending on the particular path chosen near the stationary point we conclude that (0, 0) is a saddle point. 5.4 Maxima and Minima subject to a constraint: Lagrange Multipliers There are many problems where we need to maximise or minimise a function subject to some kind of constraint. Examples are: maximise the profit for some given costs, maximise the distance travelled for a given amount of fuel, minimise the surface area for a given enclosed volume and so on. Two examples will illustrate these types of problems. Then we will discuss the method of Lagrange Multipliers when the standard approach cannot be used. Example 5.6 Minimise f(x, y) = x 2 + y 2 subject to y = x + 1. Here y = x + 1 is the constraint and so we simply substitute this constraint directly into the function. Hence, f(x, y(x)) = x 2 + ( x + 1) 2 = x 2 + x 2 2x + 1 = 2x 2 2x + 1. This is simply a function of one variable and we may find the turning point by simple differentiation. ( ) df = 4x 2 = 0, x = 1 dx 2. y= x+1 From the constraint, y = x + 1 = 1 2 + 1 = 1 2. The function evaluated at this turning point is f(1/2, 1/2) = ( 1 2 )2 + ( 1 2 )2 = 1 2. It is clear (from f (x) = 4 > 0) that this is a minimum.

76 CHAPTER 5. TAYLOR SERIES AND STATIONARY POINTS Example 5.7 Minimise f(x, y, z) = x 2 + 2y 2 + z subject to the constraint x + y 2 z = 0. To progress we substitute for z to get a function of two variables. Thus, f(x, y, z(x, y)) = f(x, y) = x 2 + 2y 2 + (x + y 2 ) = x 2 + x + 3y 2. To locate the stationary point, we evaluate the first derivatives, Therefore, = 2x + 1 = 0, x Now we calculate at ( 1/2, 0). and x = 1 2, y = 0. = f xx f yy (f xy ) 2, = 6y = 0. f xx = 2, f yy = 6, f xy = 0. Thus, = 2 6 = 12 > 0 and 2 f/ x 2 = 2 > 0, so ( 1/2, 0) is a minimum. Note that substitution is not always possible. Also the constrained functions may be a function of several variables. For such cases we need another method and use a Lagrange Multiplier that is denoted by λ. This method determines the stationary points but not their nature. Suppose we wish to find the stationary points of f(x, y, z) subject to a constraint, g(x, y, z) = 0. We define the function F (x, y, z, λ), F (x, y, z, λ) = f(x, y, z) λg(x, y, z). (5.7) λ is the Lagrange multiplier. If the constraint g(x, y, z) = 0 is satisfied, then the maximum/minimum of F will coincide with those of the constrained f. To locate the stationary points of F we need to locate the zeros of the first derivatives. Thus, x z λ = x λ g x = 0 = λ g = z λ g z = g(x, y, z) = 0 Note that these equations can also be written in a vector form as To apply the method we f = λ g, g(x, y, z) = 0.

5.4. MAXIMA AND MINIMA SUBJECT TO A CONSTRAINT: LAGRANGE MULTIPLIERS77 1. Add an unknown multiple ( λ) of the constraint g to f, then set the partial derivatives of (f λg) with respect to x, y and z to zero. This gives three equations. 2. Use the above 3 equations together with the constraint g(x, y, z) = 0 to find x, y, z and λ. The method can be generalised for more variables and more constraints. Example 5.8 Consider our first example again, where f(x, y) = x 2 +y 2 and g(x, y) = x+y 1. Define, F = x 2 + y 2 λ(x + y 1). The first derivatives become x = 2x λ, = 2y λ, λ = (x + y 1). and setting them equal to zero gives x = λ 2, y = λ 2, x + y = 1. Substituting for x and y in terms of λ in the last equation gives, Hence, x + y = 1, λ 2 + λ 2 = λ = 1. x = 1 2, y = 1 2, f = 1 4 + 1 4 = 1 2 Consider the two variable case. Why does f = λ g maximise/minimise f(x, y) along the path g(x, y) = 0? This is illustrated in Figures 5.4 and 5.5. If f, subject to g = 0 has a max/min at (x 0 ), y 0 ), then (df/ds) x0,y 0 = 0 there, where ( ) df = [( f) û] ds = 0. (x0,y 0) (x 0,y 0) Thus, û(x 0, y 0 ) is perpendicular to ( f) (x0,y 0 ). From above, û is perpendicular to g. Therefore, f and g are parallel at (x 0, y 0 ). Thus, f = λ g. Physically, following a path defined by g = 0 we would walk uphill and then downhill. However, we would walk parallel to a contour at the changeover between up and down.

78 CHAPTER 5. TAYLOR SERIES AND STATIONARY POINTS Figure 5.4: (a) Contours of constant f are shown as circles with the maximum value at f 3. (b) The path g(x, y) = 0 and the neighbouring contour g = g 1. Example 5.9 Find the stationary points of f(x, y, z) = x + y + z, on a sphere of radius 5 centred on the origin. Thus, the constraint is g(x, y, z) = x 2 + y 2 + z 2 25 = 0. The modified function with the Lagrange multiplier is F (x, y, z, λ) = f λg = x + y + z λ ( x 2 + y 2 + z 2 25 ). The first derivatives can calculated and set equal to zero to locate the stationary points. x = 1 2λx, x = 1 2λ, = 1 2λy, y = 1 2λ,

5.4. MAXIMA AND MINIMA SUBJECT TO A CONSTRAINT: LAGRANGE MULTIPLIERS79 Figure 5.5: Combining the path g = 0 with the contours of constant f. The dashed contour gives the max/min contour of the constrained f. z λ = 1 2λz, z = 1 2λ, = g = ( x 2 + y 2 + z 2 25 ) = 0. Substitute in the values for x, y and z into the last equation to get ( 1 4λ 2 + 1 4λ 2 + 1 ) ( ) 3 4λ 2 25 = 4λ 2 25 = 0. Solving for λ gives λ 2 = 3 3 100, λ = ± 10. So there are two stationary points. We have λ = + 3/10, so that (x, y, z) = (5/ 3, 5/ 3, 5/ 3) and f = 5 3. λ = 3/10, so that (x, y, z) = ( 5/ 3, 5/ 3, 5/ 3) and f = 5 3. To determine the nature of the stationary points we need to continue carefully. Setting x = 5/ 3 + x1 and y = 5/ 3 + y1, we can substitute this into the

80 CHAPTER 5. TAYLOR SERIES AND STATIONARY POINTS constraint. Then, we find that z = + 25 ( 5 3 + x1) 2 ( 5 3 + y1) 2. This may be simplified and expanded in a Taylor series of two variables for small values of x1 and y1 to give z = + 25 50 3 10 (x1 + y1) x1 2 y1 2 3 25 = + 3 10 (x1 + y1) x1 2 y1 2 3 = 5 1 2 3 3 5 (x1 + y1) 3 25 (x12 + y1 2 ) = 5 3 (1 3 5 (x1 + y1) 3 50 (x12 + y1 2 ) 1 ) 3 8 (2 5 )2 (x1 + y1) 2 +. So we may express z as the value at the stationary point plus a correction due to the value of x1 and y1. Note that we used the binomial expansion of 1 + x = 1 + x/2 x 2 /8 +. Hence, substituting x = 5/ 3 + x1 and y = 5/ 3 + y1 into f gives f = 5 + x1 + 5 + y1 5 3 (x1 + y1) 3 3 3 10 (x12 + y1 2 ) 5 8 3 (x1 + y1)2. Simplifying gives f = 5 3 3 10 (x12 + y1 2 ) 5 8 3 (x1 + y1)2. Thus, it is clear the this stationary point corresponds to a maximum. Similarly, we may show that (x, y, z) = ( 5/ 3, 5/ 3, 5/ 3) corresponds to a minimum. Example 5.10 What is the maximum area of a rectangle enclosed by the x and y axes and the curve (an ellipse) x 2 /a 2 + y 2 /b 2 = 1? This is illustrated in Figure 5.6. The area is given by f(x, Y ) = XY, and the values of X and Y are constrained by X 2 a 2 + Y 2 b 2 = 1.

5.4. MAXIMA AND MINIMA SUBJECT TO A CONSTRAINT: LAGRANGE MULTIPLIERS81 Figure 5.6: The area of the rectangle is shaded. The curve is the ellipse. So we choose Hence, g(x, Y ) = X2 a 2 + Y 2 b 2 1. ( X 2 F (X, Y, λ) = XY λ a 2 + Y 2 ) b 2 1. Now form all the first derivatives and set equal to zero. X = Y 2λX a 2 = 0, Y = 2λ a 2 X (5.8) = X 2λY Y b 2 = 0, X = 2λ b 2 Y (5.9) ( X 2 = λ a 2 + Y 2 ) b 2 1 = 0. (5.10) (5.8) and (5.9) give Y = 2λ a 2 2λ b 2 Y. Dividing by Y and solving for λ gives λ = ± ab 2. (5.11)

82 CHAPTER 5. TAYLOR SERIES AND STATIONARY POINTS (5.8) and (5.11) Y = 2 ( a 2 X ± ab ). 2 However, both X and Y must be positive and so we discard the negative sign. Thus, Y = b X. a (5.12) (5.10) and (5.12) give X 2 a 2 + 1 ( b 2 X 2 ) b 2 a 2 = 1, 2 X2 a 2 = 1, X = a. 2 (5.13) Finally, (5.12) and (5.13) give Y = b ( ) a, a 2 Y = b. 2 Therefore, the maximum area is XY = ab 2.