Maxima & Minima with Constrained Variables

BY:Arpit Modh (16BCH035)B.Tech ChemicalNirma University,Ahmedabad.

Definitions:-

Let, u = f (x , y) be a continuous function of x and y. Then u will be maximum at x = a, y = b, if f (a ,b ) > f(a + h , b + k) and will be minimum at x=a, x=b, if f(a, b) < f(a + h, b + k) for small positive or negative values of h and k.

The point at which function f(x, y) is either maximum or minimum is known as stationary point.

The value of the function at stationary point is known as extreme (maximum and minimum) value of function f(x, y).

Working Rule:-

To determine the maxima and minima (extreme values) of a function f(x, y).

Step 1: Solve ∂f/ ∂x = 0 and ∂f/ ∂y = 0 simultaneously for x and y.Step 2: Obtain the values of r= ∂²f/ ∂x², s= ∂²f / ∂x², t= ∂²f/ ∂x².

Step 3:(i) If rt - s² > 0 and r < 0 (or t < 0) at (a, b) then f(x, y) is

maximum at (a, b) and the maximum value of the function is f(a, b).

(ii) If rt - s² > 0 and r > 0 (or t > 0) at (a, b) then f(x, y) is minimum value of the function is f(a, b).

(iii) If rt - s² < 0 at (a, b) then f(x, y) is either maximum nor minimum at (a, b). Such a point is known as saddle point.

(iv) If rt - s² = 0 at (a, b) then no conclusion can be made about the extreme values of f(x, y) and further investigation is required.

Example 1Find the minimum value of x² + y² + z² with the constraint x + y + z = 3a.Solution: f = x² + y² + z²

x + y + z = 3az = 3a - x - y …..(1)

substituting the value of z in Eq. (1),f = x² + y² + (3a –x- y) ²

Step 1 For extreme values, ∂f/ ∂x = 0 and ∂f/ ∂y = 0

2x – 2(3a - x - y ) = 0 2y - 2(3a - x - y) = 04x - 6a + 2y = 0 2y - 6a + 2x + 2y = 0 2x + y = 3a x + 2y = 3a ……(2) ….(3)

Solving Eqs (2) and (3),x = y = a

The stationary point is (a, a).

Step 2 r = ∂²f/ ∂x² = 4s = ∂²f/ ∂x ∂y = 2t = ∂²f/ ∂y² = 4

Step 3 At (a, a), r = 4, s = 2, t = 4rt - s² = (4)(4) – (2) ² = 12 > 0

Also, r = 4 > 0Hence, f(x, y) is minimum at (a, a)

fmin = a² + a² + (3a - a - a) ² = 3a²

Example 2Divide 120 into three parts so that the sum of their products taken two at a time shall be maximum.Solution: Let x, y, z be three numbers.

x + y + z = 120f = xy + yz + xz = xy + y(120 - x - y) + x( 120 - x - y) = xy + 120y – xy - y² + 120x - x² -xy = 120x + 120y - xy - x² - y²

For extreme values, ∂f/∂x = 0 120 - y - 2x = 0 …(1)

And ∂f/ ∂y = 0 120 - x - 2y = 0 ….(2)

Solving Eqs (1) and (2),x = 40y = 40

Stationary point is (40, 40).

Step 2 r = ∂f/ ∂x = -2s = ∂f/ ∂x ∂y = -1t = ∂f/ ∂y = -2

Step 3 At (40, 40)rt - s² = (-2)(-2) – (-1) ² = 3 > 0r = -2 < 0

Hence, f(x, y) is maximum at (40, 40).

Example 3Find the points on the surface z²=xy+1 nearest to the origin. Also find that distance.Solution:Let p(x, y, z) be any point on the surface z² = xy + 1.Its distance from the origin is given by d² = x² + y² + z²Since p lines on the surface z² = xy + 1 d² = x² + y² + xy + 1Let f(x, y) = x² + y² + xy + 1

Step 1 For extreme values , ∂f/ ∂x=0 2x+y=o ∂f/ ∂y=0 2y+x=0Solving Eqs (1) and (2), x=0 , y=0

Step 2 r = ∂²f/ ∂x² = 2 s = ∂²f/ ∂x ∂y = 1 t = ∂² f/ ∂y² = 2

Step 3 At (0,0), r = 2, t = 2, s = 1 rt - s² = (2)(2) - 1² = 3 > 0Also, r = 2 > 0f(x, y) , i.e. d ² is minimum at (0,0) and hence d is minimum at (0,0).At (0,0), z²=xy+1=1 z=+1,z=-1 Hence, d is minimum at (0,0,1) and (0,0,-1).The points (0,0,1) and (0,0,-1) on the surface z² = xy + 1 are the nearest to the origin Minimum distance = 1.

Thank you