Maxima and Minima

 Course Outline

&    Definitions of increasing and decreasing functions, concavity, convexity, point of inflection, stationary point, saddle point.

&     Conditions for concavity and convexity.

&     Necessary and sufficient condition for maximum and minimum of functions of one, two or three variables.

&        Extreme values under subsidiary conditions.

&        Lagrange's method of undetermined multipliers

&        Problems on maxima and minima of two or three variables.

Introduction

Maximum and minimum value of a function of single variable

A function f(x) defined on some neighbourhood of 'c', then f(c) is said to be a maximum or minimum value of f(x) if there exists d > 0 such that " x Î (c – d, c + d) other than c, f(c) > f(x) or f(c) < f(x).f(c) is said to be an extreme value of f(x) if it is either a maximum or minimum value of f(x).

Maximum and minimum value of function of two and three variables

A function u = f(x, y) of two independent variables x and y is said to have the maximum or minimum value at the point (a, b) if there exist a positive number d such that for all x, y in a – d < x < a + d and b – d < y < b + d (x ¹ a, x ¹ b) satisfying
f(x, y) < f(a, b) or f(x, y) > f(a, b).

Similarly, a continuous function u = f(x, y, z) of three independent variables x, y and z is said to have maximum or minimum value at the point (a, b, c) if there exists a positive number d such that for all x, y, z in a – d < x < a + d, and b – d < y < b + d, c – d < z < c + d (x ¹ a, y ¹ b, z ¹ c). Satisfying f(x, y, z) < f(a, b, c) or f(x, y, z)

> f(a, b, c).

Stationary Points

i)       A point c is called stationary point of single variable function f(x) if
f '(c) = 0.

ii)      A point (a, b) is called stationary point of function of two variables
f(x, y) if fx (a, b) = 0 and fy (a, b) = 0.

iii)     A point (a, b, c) is said to be an extreme value of the function f(x, y, z) of three variables, if f_x (a, b, c) = 0, f_y (a, b, c) = 0 and f_z (a, b, c) = 0.

Criteria for maxima and minima.

i)       If f '(c) = 0 and f '' (c) exists then f(c) is maximum or minimum according as
f ''(c) < 0 or f '' (c) > 0.

ii)      If fx(a, b) = 0, fy (a, b) = 0 and second order partial derivatives exist in each neighbourhood of (a, b), then f(a, b) is said be

a)          Maximum value if f_xx < 0 and f_xx f_yy –(f_xy)² > 0

b)         Minimum value if f_xx > 0 and f_xx f_yy –(f_xy)² > 0

c)          Neither maximum nor minimum value if f_xx ¹ 0, and f_xx f_yy –(f_xy)² < 0.

d)         The case is doubtful if f_xx f_yy – f(_xy)² = 0.

iii)    If f_x (a, b, c) = f_y (a, b, c) = f_z (a, b, c) = 0 and all second order partial derivatives exist in neighbourhood of (a, b, c) then f(a, b, c) is said to be

Solved Examples 

Q.1.  State the condition for f(x) to have an extreme value at x = c. Find the maximum and minimum value of f(x) = 5x⁶ – 18x⁵  + 15x⁴  – 10.                      [T.U. 2075]

 Soln:    Condition for f(x) to have an extreme value at x = c: Let a function f(x) be defined at a point x = c of the interval and if f ' (c) = 0 and f '' (c) ¹ 0, then f(c) is a maximum if f '' (c) is negative and a minimum if f '' (c) is positive.

For second part:

Here, f(x) = 5x⁶ – 18x⁵ + 15x⁴ – 10

Differentiating with respect to x, we have

f ' (x) =  30x⁵ – 90x⁴ + 60x³

For maximum and minimum value of f(x),

f ' (x) = 0

or,        30x⁵ – 90x⁴ + 60x³ = 0

or,        30x³ (x² – 3x + 2) = 0

or,        x³ (x² – 2x – x + 2) = 0

or,        x³ [x (x – 2) – 1 (x – 2)] = 0

or,        x³ (x – 2) (x – 1) = 0

∴    x = 0, 1, 2 which are the stationary points.

Again, f '' (x) = 150x⁴ – 360x³ + 180x²

At x = 1, f '' (x) = 150 – 360 + 180 = – 30 < 0

∴          The function f(x) has maximum value at x = 1.

The maximum value = f(1)

= 5 – 18 + 15 – 10

= – 8

At x = 2, f '' (x) = 150 × (2)⁴ – 360 × 2³ + 180 × 2²

= 2400 – 2880 + 720

= 240 > 0

∴    f(x) is minimum at x = 2.

The minimum value = f (2)

= 5 × 2⁶ – 18 × 2⁵ + 15 × 2⁴ – 10

= 3200 – 576 + 240 – 10

= – 26

At x = 0, f '' (0) = 0. So the test fails and test by using higher order derivative.

f ''' (x) = 600x³ – 1080x² + 360x                        ∴ f ''' (0) = 0

f ^iv (x) = 1800x² – 2160x + 360                          ∴ f ^iv (0) = 360

Since the even order (fourth order) derivative is positive at x = 0, so f(x) is minimum at x = 0 and the minimum value at x = 0 is f(0) = – 10.

∴ f(x) is max. at x = 1 and max. value = –8, min. at x = 2 and min. value
= – 26, min. at x = 0 and min. value = – 10. Ans

Q.2.  Find the maximum or minimum value of the function u = x² + 3xy – 5y² subject to the constraint 2x + 3y = 6.           [T.U. 2075]

Soln :   Here, the given objective function is

u = x² + 3xy – 5y² .................. (i)

and the given constraint is

             2x+ 3y = 6 ............................ (ii)

Eliminating y between (i) and (ii), we have

Q.3.  Find the altitude of right circular cone of maximum volume that can be inscribed in a sphere of radius a.  [T.U. 2074, 2068]

Soln:    Let ABC be a right circular cone with vertex at A. Let r and h be the radius f the base and height of the cone and a be the radius of the sphere with centre at O.

             From the figure,

                          (CD)²    = (OC)² – (OD)²

                                        = a² – (DA – OA)² = a² – (h – a)²

             \          r²           = 2ah – h²

Q.4.      Show that f(x, y) = x⁴ + x²y + y² has a minimum value at origin.                 

                                                                                                                       [T.U. 2074]

Soln:    Here, f(x, y) = x⁴ + x²y + y²

             Differentiating partially, we have

                         f_x = 4x³ + 2xy,        f_y = x² + 2y

                         f_xx = 12x² + 2y       f_yy­ = 2,      f_xy = 2x

             At (0, 0), f_x = 0, f_y = 0, so (0, 0) is the stationary point of given function.

             Also, at (0, 0), f_xx = 0, f_yy = 2 and f_xy = 0

             Now, f_xx f_yy – (f_xy)² = 0.2 – 0 = 0

             Since f_xx f_yy – (f_xy)² = 0

             We get no information, so further analysis is necessary.

             Now, for small value of h and k, we have

                         f(0 + h, 0 + k) – f(0, 0) = f(h,  k) – f(0, 0)

                                                                = h⁴ + h²k + k² – 0  [∵ f(0, 0) = 0]

Q.5.      Discuss the maxima or minima of the function f(x, y) = x³ + y³ + 6xy. Also find the maximum     and minimum values.                [T.U. 2073]

Soln:    Here,    f(x, y) = x³ + y³ + 6xy

             \         f_x = 3x² + 6y,   f_y = 3y² + 6x

                         f_xx = 6x             f_xy = 6,                    f_yy = 6y

             For extreme values of f(x, y),

                         f_x = 0 and f_y = 0, therefore

                         3x² + 6y = 0 and 3y² + 6x = 0

             or,        x² + 2y = 0 and y² + 2x = 0

             Solving these equation we get

                         x = 0, y = 0 and x = – 2, y = – 2

             So the stationary points are (0, 0) and (– 2, – 2).

             At (0, 0), f_xx = 0, f_xy = 6 and f_yy = 0

             Now, f_xx= 0, f_xxf_yy – f_xy² = 0 × 0 – 6² = – 36 < 0, so f(x, y) has no maximum and minimum value at (0, 0).

             Again, at (– 2, – 2)

                         f_xx = – 12, f_xy= 6, f_yy= – 12

             Here, f_xx = – 12 < 0, f_xxf_yy – f_xy² = – 12 × – 12 – 6² = 108 > 0.

             Hence, f(x, y) has maximum value at (– 2, – 2) and the maximum value
            = f(– 2, – 2) = (– 2)³ + (– 2)³ + 6 × – 2 × – 2 = 8 Ans.

Q.8.      Find the maximum and minimum value of the function

f(x) = 5x⁶ – 18x⁵ + 15x⁴ – 10.                                                            [T.U. 2067]

Soln:    Here, f(x) = 5x⁶ – 18x⁵ + 15x⁴ – 10

             Differentiating with respect to x, we have

                          f'(x) = 30x5 – 90x4 + 60x3

                   for extreme values

                          f'(x) = 0

             \ 30x5 – 90x4 + 60x3 = 0

             or, 30x3 (x2 – 3x + 2) = 0

             or, x3 (x – 2) (x – 1) = 0

             \ x = 0, x = 1, x = 2 are stationary points.

             Again, f''(x) = 150x4 – 360x3 + 180x2

                   When x = 1, f''(x) = 150 – 360 + 180 = –30 < 0

             \ f(x) is maximum for x = 1 and the maximum value f(1) = 5 – 18 + 15 – 10 = –8

             When x = 2, f''(x)     = 150 × 24 – 360 × 23 + 180 × 22

                                                                     = 150 ×16 – 360 × 8 + 180 + 4

                                              = 2400 – 2880 + 720

                                              = 240 > 0

             \ f(x) is maximum at x = 2. The minimum value is

             f(2)             = 5 × 26 – 18 × 25 + 15 × 24 – 10

                                = 5 × 64 – 18 × 32 + 15 × 16 – 10

                                = 320 – 576 + 240 – 10

                                = – 26

             When x = 0, f''(0) = 0. The test fails and we need to examine higher order derivatives.

             f'''(x) = 600x3 – 1080x2 + 360x  \ f'''(0) = 0

             fiv(x) = 1800x3 – 2160x + 360  \ fiv(0) = 360 > 0

             Since even order derivative is positive for x = 0 and f(x) minimum for
x = 0. The minimum value at x = 0 is f(0) = – 10.