JUST THE MATHS SLIDES NUMBER 14.11 PARTIAL DIFFERENTIATION 11 - PDF document

“JUST THE MATHS” SLIDES NUMBER 14.11 PARTIAL DIFFERENTIATION 11 (Constrained maxima and minima) by A.J.Hobson 14.11.1 The substitution method 14.11.2 The method of Lagrange multipliers

UNIT 14.11 PARTIAL DIFFERENTIATION 11 CONSTRAINED MAXIMA AND MINIMA We consider the determination of local maxima and lo- cal minima for a function, f ( x, y, ... ), subject to an addi- tional constraint in the form of a relationship, g ( x, y, ... ) = 0. This would occur, for example, if we wished to construct a container with the largest possible volume for a fixed value of the surface area. 14.11.1 THE SUBSTITUTION METHOD The following examples illustrate a technique for elemen- tary cases: EXAMPLES 1. Determine any local maxima or local minima of the function, f ( x, y ) ≡ 3 x 2 + 2 y 2 , subject to the constraint that x + 2 y − 1 = 0. Solution Here, it is possible to eliminate either x or y by using the constraint. 1

If we eliminate x , we may write f ( x, y ) as a function, F ( y ), of y only. f ( x, y ) ≡ F ( y ) ≡ 3(1 − 2 y ) 2 + 2 y 2 ≡ 3 − 12 y + 14 y 2 . Using the principles of maxima and minima for func- tions of a single independent variable, F ′ ( y ) ≡ 28 y − 12 and F ′′ ≡ 28 . A local minimum occurs when y = 3 / 7 and, hence, x = 1 / 7. The corresponding local minimum value of f ( x, y ) is 2 2  1  3 = 21 49 = 3     3 + 2 7 .       7 7 2. Determine any local maxima or local minima of the function, f ( x, y, z ) ≡ x 2 + y 2 + z 2 , subject to the constraint that x + 2 y + 3 z = 1. Solution Eliminating x , we may write f ( x, y, z ) as a function, F ( y, z ), of y and z only. f ( x, y, z ) ≡ F ( y, z ) ≡ (1 − 2 y − 3 z ) 2 + y 2 + z 2 . 2

That is, F ( y, z ) ≡ 1 − 4 y − 6 z + 12 yz + 5 y 2 + 10 z 2 . Using the principles of maxima and minima for func- tions of two independent variables, ∂F ∂y ≡ − 4 + 12 z + 10 y and ∂F ∂z ≡ − 6 + 12 y + 20 z. A stationary value will occur when these are both equal to zero. Thus, 5 y + 6 z = 2 , 6 y + 10 z = 3 , which give y = 1 / 7 and z = 3 / 14. The corresponding value of x is 1 / 14, which gives a stationary value for f ( x, y, z ) of 14 / (14) 2 = 1 14 . Also, ∂ 2 F ∂ 2 F ∂ 2 F ∂y 2 ≡ 10 > 0 , ∂z 2 ≡ 20 > 0 , and ∂y∂z ≡ 12 . Thus, 2 ∂ 2 F ∂y 2 .∂ 2 F  ∂ 2 F   = 200 − 144 > 0 . ∂z 2 −     ∂y∂z  3

Hence there is a local minimum value, 1 14 , of x 2 + y 2 + z 2 , subject to the constraint that x + 2 y + 3 z = 1, at the point where x = 1 14 , y = 1 7 , and z = 3 14 . Note: Geometrically, this example is calculating the square of the shortest distance from the origin onto the plane whose equation is x + 2 y + 3 z = 1. 14.11.2 THE METHOD OF LAGRANGE MULTIPLIERS In determining the local maxima and local minima of a function, f ( x, y, ... ) subject to the constraint that g ( x, y, ... ) = 0, it may be inconvenient (or even impossible) to elimi- nate one of the variables, x, y, ... . The following example illustrates an alternative method for a function of two independent variables: (a) Suppose that the function z ≡ f ( x, y ) is subject to the constraint that g ( x, y ) = 0. Then, z is effectively a function of x only. 4

The stationary values of z will be determined by the equa- tion, d z d x = 0 . (b) The total derivative of z ≡ f ( x, y ) with respect to x , when x and y are not independent of each other, is given by d z d x = ∂f ∂x + ∂f d y d x. ∂y (c) From g ( x, y ) = 0, the process used in (b) gives ∂x + ∂g ∂g d y d x = 0 . ∂y Hence, for all points on the surface with equation, g ( x, y ) = 0, ∂g d y ∂x d x = − . ∂g ∂y 5

Thus, throughout the surface with equation, g ( x, y ) = 0, � ∂g � d z d x = ∂f  ∂f   ∂x � . ∂x −     � ∂g ∂y  ∂y (d) Stationary values of z , subject to the constraint that g ( x, y ) = 0, will occur when ∂f ∂x.∂g ∂y − ∂f ∂y.∂g ∂x = 0 . This may be interpreted as the condition that the two equations, ∂f ∂x + λ∂g ∂x = 0 and ∂f ∂y + λ∂g ∂y = 0 , should have a common solution for λ . (e) Suppose that φ ( x, y, λ ) ≡ f ( x, y ) + λg ( x, y ) . Then, φ ( x, y, λ ) would have stationary values whenever its first order partial derivatives with respect to x , y and λ were equal to zero. 6

In other words, ∂f ∂x + λ∂g ∂f ∂y + λ∂g ∂x = 0 , ∂y = 0 , and g ( x, y ) = 0 . Conclusion The stationary values of the function z ≡ f ( x, y ), subject to the constraint that g ( x, y ) = 0, occur at the points for which the function, φ ( x, y, λ ) ≡ f ( x, y ) + λg ( x, y ) , has stationary values. The number, λ , is called a “Lagrange multiplier” Notes: (i) To determine the nature of the stationary values of z , it will usually be necessary to examine the geometrical conditions in the neighbourhood of the stationary points. (ii) The Lagrange multiplier method may also be applied to functions of three or more independent variables. 7

EXAMPLES 1. Determine any local maxima or local minima of the function, z ≡ 3 x 2 +2 y 2 , subject to the constraint that x + 2 y − 1 = 0. Solution Firstly, we write φ ( x, y, λ ) ≡ 3 x 2 + 2 y 2 + λ ( x + 2 y − 1) . Then, ∂φ ∂x ≡ 6 x + λ, ∂φ ∂y ≡ 4 y + 2 λ and ∂φ ∂λ ≡ x + 2 y − 1 . The third of these is already equal to zero; but we equate the first two to zero, giving 6 x + λ = 0 , 2 y + λ = 0 . Eliminating λ shows that 6 x − 2 y = 0, or y = 3 x . Substituting this into the constraint, 7 x − 1 = 0. Hence , x = 1 7 , y = 3 7 and λ = − 6 7 . 8

A single stationary point occurs at the point where 2 2 x = 1 7 , y = 3  1  3 = 3     7 and z = 3 + 2 7 .       7 7 The geometrical conditions imply that the stationary value of z occurs at a point on the straight line whose equation is x + 2 y − 1 = 0. y ❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍ ✻ ✲ x O The stationary point is, in fact, a minimum value of z since the function, 3 x 2 + 2 y 2 , has values larger than 3 / 7 ≃ 0 . 429 at any point either side of the point, (1 / 7 , 3 / 7) = (0 . 14 , 0 . 43), on the line with equation, x + 2 y − 1 = 0. For example, at the points 0 . 12 , 0 . 44) and (0 . 16 , 0 . 42) on the line, the values of z are 0 . 4304 and 0 . 4296, respectively. 2. Determine the maximum and minimum values of the function, z ≡ 3 x + 4 y , subject to the constraint that x 2 + y 2 = 1. 9

Solution Firstly, we write φ ( x, y, λ ) ≡ 3 x + 4 y + λ ( x 2 + y 2 − 1) . Then, ∂φ ∂x ≡ 3+2 λx, ∂φ ∂y ≡ 4+2 λy and ∂φ ∂λ ≡ x 2 + y 2 − 1 . The third of these is already equal to zero; but we equate the first two to zero, giving 3 + 2 λx = 0 , 2 + λy = 0 . Thus, x = − 3 2 λ and y = 2 λ. Substituting into the constraint gives 4 λ 2 + 4 9 λ 2 = 1 . That is, 9 + 16 = 4 λ 2 and hence λ = ± 5 2 . Hence, x = ± 3 5 and y = ± 4 5 , giving stationary values, ± 5, of z . 10

Finally, the geometrical conditions suggest that we consider a straight line with equation, 3 x +4 y = c , (a constant) moving across the circle with equation, x 2 + y 2 = 1. ❙ y ❙ ✻ ❙ ❙ ❙ ❙ ❙ ❙ ✲ x ❙ O ❙ ❙ ❙ ❙ ❙ ❙ ❙ ❙ ❙ ❙ ❙ ❙ ❙ ❙ ❙ The further the straight line is from the origin, the greater is the value of the constant, c . The maximum and minimum values of 3 x +4 y , subject to the constraint that x 2 + y 2 = 1, will occur where the straight line touches the circle. We have shown that these are the points, (3 / 5 , 4 / 5) and ( − 3 / 5 , − 4 / 5). 3. Determine any local maxima or local minima of the function, w ≡ x 2 + y 2 + z 2 , subject to the constraint that x + 2 y + 3 z = 1. 11

Solution Firstly, we write φ ( x, y, z, λ ) ≡ x 2 + y 2 + z 2 + λ ( x + 2 y + 3 z − 1) . Then, ∂φ ∂φ ∂φ ∂x ≡ 2 x + λ, ∂y ≡ 2 y + 2 λ, ∂z ≡ 2 z + 3 λ and ∂φ ∂λ ≡ x + 2 y + 3 z − 1 . The fourth of these is already equal to zero; but we equate the first three to zero, giving 2 x + λ = 0 , y + λ = 0 , 2 z + 3 λ = 0 . Eliminating λ shows that 2 x − y = 0, or y = 2 x , and 6 x − 2 z = 0, or z = 3 x . Substituting these into the constraint gives 14 x = 1. Hence, x = 1 14 , y = 1 7 , z = 3 14 and λ = − 1 7 . 12

A single stationary point occurs, therefore, at the point where x = 1 14 , y = 1 7 , z = 3 14 and 2 2 2  1  1  3 = 1       w = + + 14 .          14 7 14 Finally the geometrical conditions imply that the sta- tionary value of w occurs at a point on the plane whose equation is x + 2 y + 3 z = 1. z ✻ �❍❍❍❍❍ � � ❍ ✲ y ✘ ✘✘✘✘✘✘✘✘✘✘✘ ✟ O � ✟ ✟ � ✟ ✟ � ✟ ✟ ✟ ✟ ✟ ✟ ✙ ✟ x The stationary point must give a minimum value of w since the function, x 2 + y 2 + z 2 , represents the square of the distance of a point ( x, y, z ) from the origin. If the point is constrained to lie on a plane, this dis- tance is bound to have a minimum value. 13