Maxima and Minima without Constraints

MAXIMA AND MINIMA WITHOUT CONSTRAINTS

 

To optimize something means to maximize or minimize some aspects of it. An important application of multivariable differential calculus is finding the maximum and minimum values of functions of several variables and determination of function of several variables and determination where they occur .In the study of stability of equilibrium states of mechanical and physical systems, determination of extreme values of constrained functions in economics, in designing multistage rockets in engineering, in geometry etc.

Geometrically z = f(x, y) respect to a surface. The maximum is a point on the surface from which the surface descends (come down) in every direction backwards xy‒plane. The minimum is the bottom direction from which the surface ascends(climbs up). In either case, the tangent planes to the surface at a maximum point is horizontal to z‒axis (parallel to the xy‒plane) and perpendicular to z‒axis.

 

 

1. MAXIMA AND MINIMA OF FUNCTIONS OF TWO VARIABLES:

 

A function f(x, y) is said to have a maximum value of x = a, y = b if f(a, b) > f(a+h, b+ k), for small and independent values of h and k, positive or negative.

A function f(x,y) is said to have a minimum value of x = a, y = b if f(a,b) < f(a + h, b + k), for small and independent values of h and k, positive or negative.

The point (a, b) is called a stationary point if f_x(a, b) = 0, f_y(a, b) = 0. The value f(a, b) is called a stationary value. Thus every extreme value is a stationary value but the converse may not be true.

A stationary value (a, b) is said to be a saddle point of (x, y) if f(x, y) attends neither maximum nor minimum at (a, b).

 

RULE TO FIND THE EXTREME VALUES OF A FUNCTION z = f(x,y)

 

1) Find f_x, f_y, f_xx, f_xy, f_yy.

 

2) Solve f_x = 0, f_y = 0. Let (a, b), (c, d) ...... be the solutions of these equations.

 

3) For each solution in steps(2), find r = f_xx, s=f_xy, t = f_yy. Let Δ = rt ‒ s².

 

4) If Δ> 0 and r<0 for a particular solution (a, b) of step (2), then f(x, y) has a maximum value at (a, b).

 

5) If Δ > 0 and r> 0 for a particular solution (a, b) of step (2), then f(x,y) has a minimum value at (a, b).

 

6) If Δ <0 for a particular solution (a, b) of step (2), then f(x, y) has no extreme value (a, b). The stationary value (a, b) is called saddle point of f(x, y).

 

7) If Δ = 0 or Δ > 0 and r = 0 or Δ > 0 and s = 0, the cases are doubtful and requires further investigation.

 

 

PROBLEMS UNDER MAXIMA AND MINIMA

 

Example 43. Examine the extreme values for x² + y²+ 6x + 12.

Solution: Given f(x, y) = x² + y² + 6x + 12

f_x = 2x+6; f_y = 2y

r = f_xx = 2; s = f_xy = 0; t = f_yy = 2

To find stationary points of f(x, y):

f_x = 0⇒2x +6= 0

x +3=0

x =‒3

f_y=0⇒ 2y = 0

y = 0

∴ Stationary point is (‒3,0).

To check for optimality:

At (‒3,0):

r=2>0

s=0

t = 2

Δ = rt‒s² = 2(2)‒0=4>0

Since r > 0 and Δ> 0, f(x, y) has minimum value at (‒3,0).

Minimum value of the function is f(‒3, 0) = 9+0‒18 + 12 = 3.

 

Example 44. Examine for extreme values of f(x, y) = x² ‒ xy + y² − 2x + y

Solution: Given f(x, y) = x² ‒ xy + y² ‒ 2x + y

f_x= 2x‒y‒2;

f_y= −x + 2y + 1

r = f_xx = 2;

s = f_xy = ‒1;

t = f_yy = 2

To find stationary points of f(x, y):

f_x = 0

2x‒y‒2=0           ….(1)

f_y = 0

− x + 2y + 1 = 0 ...... (2)

(2) × 2 ⇒ −2x + 4y + 2 = 0       …… (3)

(1) + (3) ⇒ 3y = 0

y = 0

Sub y = 0 in (1) → 2x‒0‒2=0

2x=2

x=1

∴ The stationary point is (1,0).

To check for optimality:

At (1,0):

r = 2 > 0

s = ‒1

t=2

Δ = rt‒s² = 2(2)‒1=3> 0

Since r > 0 and Δ > 0, f(x, y) has minimum value at (1, 0).

Minimum value of the function is f(1, 0) = 1‒0+0‒2+0 = ‒1.

 

Example 45. Prove that f(x, y) = 2x²y + x² − y² + 2y has no extremum.

Solution: Given f(x, y) = 2x²y + x²‒y²+2y

f_x = 4xy + 2x, f_y = 2x² ‒ 2y+2,

r = f_xx = 4y+2, s= f_xy = 4x, t = f_yy=‒2

To find stationary points of f(x, y):

f_x = 0⇒ 4xy + 2x = 0

2x(2y+ 1) = 0

f_y = 0⇒ 2x² ‒ 2y + 2 = 0

x² ‒ y + 1 = 0 ... (1)

2x = 0 or y = ‒1/2

When x = 0, (1) ⇒ y = 1

The point is (0,1)

When y = ‒1/2, (1) ⇒ x = √(‒3/2) = i√(3/2) which is imaginary

.. Stationary point is (0,1)

To check for optimality

At (0, 1):

r = 4y+2 = 6 > 0

s = 4x = 0

t = ‒2

Δ = rt‒s² = 6(‒2)‒0 = ‒12 < 0

Since Δ < 0, (0,1) is a saddle point.

..Given function has no extreme value.

 

Example 46. Examine f (x, y) = x³ + y³ ‒ 12x‒3y+ 20 for extreme values.

Solution: Given: f(x, y) = x³+ y³‒12x‒3y+20

f_x = 3x² ‒ 12;

f_y = 3y²‒3

 r = f_xx = 6x;  s = f_xy = 0; t = f_yy=6y

To find stationary points of f(x, y):

fx = 0

3x²‒12=0

3x² = 12

x² = 4

x = +2

f_y = 0

3y² ‒ 3 = 0

3y² = 3

y²=1

y = +1

∴ The stationary points are (2,1), (2,‒1), (‒2,1), (‒2‒1)

To check for optimality:

At (2, 1):

r = 6x = 6(2) = 12 > 0

s=0

t = 6y = 6(1) = 6

Δ = rt ‒ s² = 12(6) ‒ 0 = 72 > 0

Since r > 0 and Δ > 0, f(x, y) has minimum value at (2, 1).

Minimum value of the function is f(2, 1) = 8+1‒24‒3+20= 2.

At (2,‒1):

 r = 6x=6(2)= 12 > 0

s=0

t = 6y=6(‒1) = ‒6

Δ = rt‒s² = 12(‒6)‒0 = ‒72 < 0

Since Δ<0, (2,‒1) is a saddle point.

At (‒2,‒1):

r = 6x = 6(‒2) = ‒12 < 0

s = 0

t = 6y = 6(−1) = −6

Δ = rt‒s² = (‒12)(‒6)‒0 = 72> 0

Since r < 0 and Δ > 0, f(x,y) has maximum value at (‒2,‒1).

Maximum value of the function is ƒ(‒2,‒1) = −8‒1+24+3+20 = 38.

At (‒2,1):

r = 6x = 6(‒2) = ‒12 < 0

s=0

t = 6y = 6(1) = 6

Δ = rt ‒ s² = (‒12)(6) — 0 = −72 < 0

Since Δ < 0, (‒2, 1) is a saddle point.

 

Example 47. Investigate the extreme values of the function f(x, y) = x² + xy + y² + 1/x + 1/y.

Solution:

Given f(x, y) = x² + xy + y² + 1/x + 1/y

f_x = 2x + y ‒ 1/x²;

f_y = x + 2y ‒ 1/y²

r = f_xx = 2 + 2/x³;

s = f_xy = 1;

t = f_yy = 2 + 3/y³

To find stationary points of f(x, y):

f_x=0

2x + y −1/x² = 0        …..(1)

f_y = 0

x + 2y −1/y² =0          ...... (2)

 (1)− (2) ⇒ x − y – 1/x² + 1/y² = 0

 x ‒ y + (x² ‒ y²)/(x²y²) = 0

 x ‒ y + [(x + y)(x − y)]/[x²y²] = 0

(x − y) ( 1 + [(x+y)/x²y²] ) = 0

x‒y=0

⇒ y=x

(1) ⇒ 2x + x ‒ 1/x² = 0

3x = 1/x²

x³ = 1/3

x = (1/3)^1/3

∴ y = (1/3)^1/3

The stationary point is (  (1/3)^1/3 ,  (1/3)^1/3 )

To check for optimality:

At (  (1/3)^1/3 ,  (1/3)^1/3 ):

 r = 2 + 2/x³ = 2+6

 = 8>0

s=1

t = f_yy = 2 + 2/y³ = 8

Δ = rt ‒ s² = 64‒1 = 63 > 0

Since r > 0 and Δ > 0, f(x, y) has minimum value at (  (1/3)^1/3 ,  (1/3)^1/3 ).

Minimum value of the function is

 f (  (1/3)^1/3 ,  (1/3)^1/3 ) = (1/3)^2/3 + (1/3)^2/3 + (1/3)^2/3 + 3^1/3 + 3^1/3

= 3(1/3)^2/3 + 2(3^1/3)

= 3(3)^-2/3 + 2(3^1/3)

= 3^1-2/3 + 2(3^1/3)

= 3^1/3 + 2(3^1/3)

= 3(3^1/3)

= 3^{1 + 2/3}

= 3^1/3

Minimum point is (  (1/3)^1/3 ,  (1/3)^1/3 ) and

The minimum value is 3^4/3.

 

Example 48. Examine for extreme values of (x, y) if f(x, y) = x³ + 3xy² −15x² – 15y² + 72x.

Solution: Given f(x, y) = x³ + 3xy² ‒ 15x²‒15y² + 72x

f_x = 3x² + 3y² ‒ 30x + 72 ;

f_y = 6xy ‒ 30y

 r = f_xx = 6x − 30;

s = f_xy = 6y;

t = f_yy = 6x ‒ 30

To find stationary points of f(x, y):

f_x = 0

3x² + 3y² ‒ 30x + 72 = 0

x²+ y² ‒ 10x + 24 = 0          …….(1)

f_y = 0

6xy ‒ 30y=0

6y(x‒5)=0

y=0, x‒5=0

Case (i) When y = 0

(1) ⇒ x² ‒ 10x + 24 = 0

(x‒4)(x‒6)=0

x=4, 6

The points are (4,0), (6,0)

Case (ii) When x = 5,

(1) ⇒ 25+ y² − 50+ 24 = 0

y² − 1 = 0 

⇒ y² = 1

y= ±1

⇒ y= ‒1, 1

The points are (5,−1), (5, 1)

.. The stationary points are (4,0), (6,0), (5,‒1), (5,1)

To check for optimality:

At (4,0):

r=6x‒30=6(4)‒30‒6 < 0

s = 6y = 6(0) = 0

t=6x‒30=6(4) − 30 = −6

Δ = rt‒s² = (‒6)(‒6) ‒ 0 = 36> 0

Since r < 0 and Δ > 0, f(x, y) has maximum value at (4,0)

Maximum value of the function = f(4,0) = 64+0‒240‒0+288 = 112.

At (6,0):

r = 6x − 30 = 6(6) — 30 = 6> 0

s = 6y=6(0)=0

t = 6x‒30 = 6(6) — 30 = 6

Δ = rt ‒ s² = (6)(6) − 0 = 36 > 0

Since r > 0 and Δ > 0, f(x, y) has minimum value at (6,0).

Minimum value of the function is f(6, 0) = 216+ 0 − 540 ‒ 0 + 432 = 108.

At (5,1):

r=6x‒30=6(5)‒30=0

s = 6y=6(1) = 6

t = 6x ‒ 30 = 6(5) ‒ 30 = 0

Δ = rt‒s² = (0) (0) ‒ 36 = −36 < 0

Since Δ < 0, (5, 1) is a saddle point.

At (5,‒1):

r=6x‒30=6(5)‒30=0

s = 6y=6(‒1) = −6

t=6x‒30=6(5) ‒ 30 = 0

Δ = rt‒s² = (0)(0) ‒ 36=‒36 < 0

Since Δ < 0, (5,‒1) is a saddle point.

 

Example 49. Examine the extreme of the function f(x, y) = x⁴ + x²y + y² at the origin.

Solution: Given f(x, y) = x² + x²y + y²

f_x = 4x³ + 2xy; f_y = x² + 2y

 r = f_xx = 12x² + 2y ;

s = f_xy = 2x;

t = f_yy = 2

The origin (0,0) satisfies f_x= 0 and f_y = 0.

∴ (0,0) is a stationary point of f(x, y).

To check for optimality:

At (0,0):

r = f_xx = 0; s = f_xy = 0;

t = f_yy = 2

Δ = rt ‒ s² = 0

Since Δ = 0, further investigation is required to find the nature of the extreme of f(x, y) at the origin. Let us consider the values of f(x, y) at the neighborhood of (0, 0) namely (h, 0), (0, k), (h, h), where ‒1 <h< 1; −1 < k < 1.

ƒ(0,0) = 0

f(0,k) = k² > 0

f(h, 0) = h⁴ > 0

f(h,h) = h⁴ + h³ + h²

= h²(1+h+h²) > 0

Thus f(x,y)>f(0,0) in the neighbour hood of (0,0).

∴ (0,0) is a minimum value of f (x,y) and the minimum value of f(x, y) = 0

 

Example 50. Locate the stationary points of x⁴ + y⁴ ‒ 2x² + 4xy – 2y².

Solution: Given f(x, y) = x⁴ + y⁴ ‒ 2x² + 4xy ‒ 2y² 

f_x = 4x³‒4x+4y;

f_y = 4y³ + 4x − 4y;

r = f_xx = 12x² ‒ 4 ;

s = f_xy = 4;

t = f_yy = 12y²‒4

To find stationary points of f(x, y):

f_x = 0

4x³‒4x+4y = 0

x³‒x+y=0       ....... (1)

f_y = 0

4y³+4x‒4y=0

y³ + x ‒ y = 0 ...... (2)

(1)+(2) ⇒x³ + y³ = 0

(x+y)(x² ‒ xy + y²) = 0

x+y=0 or x²‒xy + y² = 0

y = −x, x² ‒ xy + y² = 0 gives complex values.

Now put y = ‒x in (1)

(1) ⇒ x³‒ x − x = 0

x³‒ 2x = 0

x(x²‒2) = 0

x = 0, x²‒2=0 ⇒ x² = 2.

x= ±√2 ⇒ x = ‒√2,√2

x = ‒√2, 0, √2

When x = 0, y = ‒x ⇒ 0

When x = √2, y = −√2

When x = ‒√2, y = √2

∴ The stationary points are (0,0), (√2, −√2 ), (−√2, √2).

To check for optimality:

At (0,0):

r = 12x² ‒ 4 = 12(0) – 4 = −4 < 0

 s = 4

 t = 12y² ‒ 4 = 12(0)  − 4 = −4

∆ = rt ‒ s² = (‒4)(‒4) ‒ 16 = 0

Since ∆ = 0, further investigation is required to find the nature of the extreme of f(x,y) at the origin. Let us consider the values of f(x, y) at the neighborhood of (0, 0) namely: (h, 0), (0, k), (h, h),where ‒1 < h < 1; −1 < k < 1.

f(0,0) = 0

f(0,k) = k²(k²‒2) < 0

f(h, 0) = h²(h²‒2) < 0

f(h,h) = 2h⁴ > 0

Thus in the neighborhood of (0,0) there are points where f(x, y) < f(0,0) and f(x,y) > f(0,0).

Hence ƒ(0,0) is not an extreme value.

∴ (0,0) is a saddle point.

At (√2,‒√2):

r = 12x² ‒ 4 = 12(2)‒4=20 > 0

 s = 4

t = 12y² ‒ 4 = 12(2)‒4 = 20

∆ = rt‒s² = (20)(20) ‒ 16 = 384 > 0

Since r > 0 and ∆ > 0, f(x, y) has a minimum value at (√2, ‒√2).

Minimum value of the function = f(√2,‒√2) = 4+4‒4‒8‒4 = ‒8.

At (−√2,√2):

 r = 12x² ‒ 4 = 12(2)‒4 = 20 > 0

s = 4

t = 12y² ‒ 4 = 12(2) ‒ 4 = 20

∆ = rt‒s² = (20)(20) ‒ 16 = 384 > 0

Since r > 0 and ∆ > 0, f(x, y) has a minimum value at (‒√2, √2).

Minimum value of the function is f(‒√2, √2) = 4 + 4 − 4 ‒ 8 ‒ 4 = −8.

∴ (√2, −√2) and (−√2, √2) are the two minimum points for the given function.

 

Example 51. Examine the function x³+ y³ ‒ 3axy for maxima and minima.

Solution:

 Given f = x³ + y³ ‒ 3axy

f_x = 3x² ‒ 3ay;

f_y=3y² ‒ 3ax

r = f_xx = 6x ;

s = f_xy = −3a ;

t = f_yy = 6y

To find'stationary points of f(x, y):

f_x = 0

3x² ‒ 3ay = 0 ... (1)

f_y = 0

3y²‒3ax = 0 ... (2)

(1)⇒3x² = 3ay

y = x²/a       ……..(3)

Substitute the values of y in (2)

(2) ⇒ 3 (x²/a)² ‒ 3ax = 0

3(x⁴/a²) ‒ 3ax = 0

[ 3x⁴ ‒ 3a³x ] / a² = 0

3x(x³‒a³) = 0

x = 0 or x³‒a³ = 0

x = 0 or x³ = a³ i.e.; x=a

x=0; x = a

When x = 0, (3) ⇒ y = 0

When x = a, (3) ⇒ y = a

Stationary points are (0,0), (a, a).

To check for optimality:

At (0,0):

r = 6x = 0

s = ‒3a

t = 6y= 0

A = rt‒s² = (0)(0) ‒ 9a² = ‒9a² < 0

Since ∆ < 0, (0,0) is a saddle point.

At (a,a):

r = 6x = 6a

s = ‒3a

t = 6y= 6a

∆ = rt ‒ s² = (6a)(6a) ‒ 9a² = 27a² > 0

Thus if a > 0, r = 6a > 0 and ∆ > 0, f(x, y) has a minimum value at (a, a) and if a < 0, r = 6a <0 and ∆ > 0, f(x, y) has a maximum value at (a, a).

 

Example 52. Locate the stationary points of x³y²(1‒x‒y) and determine their nature.

Solution:

Given f(x, y) = x³y²(1‒x‒y)

= x³y²‒x⁴y² ‒ x³y³

 f_x = ∂f/∂x = 3x²y² ‒ 4x³y² ‒ 3x²y³

f_y = ∂f/∂y = 2x³y‒2x⁴y‒ 3x³y²

r = f_xx = a²f/ax² = 6xy² ‒ 12x²y² ‒ 6xy³

s = f_xy = a²f/მy² = 6x²y‒8x³y‒9x²y²

t = f_yy = a²f/მy² = 2x³ ‒ 2x⁴ ‒ 6x³y

To find stationary points of f(x, y):

 f_x = 0

 3x²y² ‒ 4x³y²‒3x²y³ = 0

x²y²(3‒4x‒3y) = 0

 x = 0, y = 0, 3‒4x‒3y = 0         ………….(1)

f_y = 0 ⇒ 2x³y‒2x⁴y‒3x³y² = 0

x³y(2‒2x‒3y) = 0

x = 0, y = 0, 2‒2x‒3y=0       ...... (2)

(1) − (2) ⇒ x = 1/2

Put x = ½ in (2), we get y = 1/3

point is (1/2,1/3)

∴. Stationary points are (1/2, 1/3) and (0,0)

At (1/2, 1/3):

 r = 6xy² – 12x²y² – 6xy³ = 6(1/2)(1/3)² ‒ 12(1/2)²(1/3)² ‒ 6(1/2)(1/3)³

= ‒1/9 < 0

 s = 6x²y – 8x³y – 9x²y² = 6(1/2)²(1/3) ‒ 8(1/2)³(1/3) ‒ 9(1/2)²(1/3)² = 1/12

 t = 2x³ ‒ 2x⁴ ‒ 6x³y = 2(1/2)³ ‒ 2(1/2)⁴ − 6(1/2)³(1/3) = ‒ 1/8

∆ = rt ‒ s² = (‒1/9)(1/8) ‒ (1/12)² = 1/144 > 0

Since r < 0 and ∆ > 0, f(x, y) has a maximum value at ( ½ , 1/3)

At (0,0):

r = 0, s = 0, t = 0

∆A = rt ‒ s² = 0

Since ∆= 0, further investigation is required to find the nature of the extreme of f(x,y) at the origin. Let us consider the values of f(x, y) at the neighborhood of (0,0) namely (h, 0), (0, k), (h, h),where ‒1 <h< 1; ‒1 < k < 1.

ƒ(0,0) = 0..

f(0,k) = 0

f(h, 0) = 0

f(h,h) = h⁵(1‒2h)

f(0.1,0.1) = (0.1)⁵(1‒2(0.1)) = (0.1)⁵(0.8) > 0

f(‒0.1, −0.1) = (‒0.1)⁵(1‒2(‒0.1)) = ‒(0.1)⁵(1.2) < 0

Thus in the neighborhood of (0,0) there are points where f(x,y) < ƒ(0,0) and f(x,y) > f(0,0).

Hence f(0,0) is not an extreme value, (0, 0) is a saddle point.

 

Example 53. Examine the extreme values of x³y²(12 – 3x – 4y).

Solution:

Given f(x, y) = x³y²(12‒3x‒4y)

f(x,y) = x³y²(12‒3x‒4y) = 12x³y² ‒ 3x⁴y² ‒ 4x³y³

f_x = 36x²y²‒12x³y² ‒ 12x²y³

f_y = 24x³y‒6x⁴y‒12x³y²

 r = f_xx = 72xy² ‒ 36x²y²‒24xy³

s = f_xy = 72x²y ‒ 24x³y ‒ 36x²y²

t = f_yy= 24x³‒6x²‒24x³y

To find stationary points of f(x, y)

f_x=0 ⇒ 36x²y² ‒ 12x³y² ‒ 12x²y³ = 0

36x²y²‒12x³y²‒12x²y³ = 0

12x²y²(3‒x‒y)=0

⇒ x²y²(3 ‒ x − y) = 0

x²y² = 0, (3‒x‒y) = 0

x = 0, y = 0, x+y=3             ………..(1)

f_y=0

⇒ 24x³y‒ 6x⁴y‒12x³y² = 0

⇒ 4x³y‒x⁴y‒2x³y² = 0

x³y(4‒x‒2y) = 0

x³y=0x+2y= 4      …….(2)

x = 0, y = 0, x + 2y = 4

(1)‒(2)=>

x +   y = 3

x + 2y = 4

  ‒y     = ‒1

Substitute y = 1, in (1), we get x + 1 = 3

⇒ x = 2

The stationary points are (2,1), (0,0)

At (2,1):

r = 72xy² ‒ 36x²y²‒24xy³ = 144‒144‒48 = ‒48 < 0

s = 72x²y ‒ 24x³y ‒ 36x²y² = 288 – 192 – 144 = −48

t = 24x³‒ 6x⁴ ‒ 24x³y = 92‒96 – 192 = ‒96

 ∆ = rt‒s² = 4608‒2304 = 2304 > 0

Since r < 0 and ∆ > 0, f(x, y) has a maximum value at (2,1)

Maximum value of the function is f(2, 1) = 16.

At (0,0): r = 0, s = 0, t = 0

 ∆ = rt ‒ s² = 0

Since ∆ = 0, further investigation is required to find the nature of the extreme of f(x,y) at the origin. Let us consider the values of f(x, y) at the neighborhood of (0,0) namely (h, 0), (0, k), (h, h),where ‒1 < h < 1; −1 < k < 1.

f(0,0) = 0

f(0,k) = 0

f(h, 0) = 0

f(h,h) = h⁵(12‒7h)

ƒ(0.1,0.1)=(0.1)⁵(12‒7(0.1)) = (0.1)⁵(11.3) > 0

ƒ(‒0.1,‒0.1) = (‒0.1)⁵(12‒7(‒0.1)) = (0.1)⁵(12.7) < 0

Thus in the neighborhood of (0, 0) there are points where f(x,y) < f(0,0) and

f(x, y) > f(0,0).

Hence f(0,0) is not an extreme value.

(0,0) is a saddle point.

 

2. LAGRANGE'S METHOD OF UNDETERMINED MULTIPLIERS

 

To find the maximum or minimum values of a function of three (or more) variables, when the variables are not independent but are connected by some given relation, we try to convert the given function to the one, having least number of independent variables with the help of the give relation. When this procedure is not practicable, we use Lagrange's method.

Let f (x, y, z) be a function of x, y, z which is to be examined for maximum or minimum value. Let the variables x, y, z be connected by the relation ϕ(x, y, z) = 0. Let u = f(x, y, z) + λϕ(x, y, z) be the auxillary function, where λ is Lagrangian multiplier.

Let

∂u/∂x=0,

∂u/∂y=0,

∂u/∂z=0,

∂u/∂λ=0.

These equations give the values of x, y, z and λ for a maximum or minimum.

Note: Lagrange's method does not enable us to find whether there is a maximum or minimum. This fact is determined from the physical considerations of the problem.