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### Maxima & Minima of Calculus

• 1. Maxima & Minima with Constrained Variables BY: Arpit Modh (16BCH035) B.Tech Chemical Nirma University, Ahmedabad.
• 2. 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).
• 3. 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.
• 4. 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.
• 5. Example 1 Find the minimum value of x² + y² + z² with the constraint x + y + z = 3a. Solution: f = x² + y² + z² x + y + z = 3a z = 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) = 0 4x - 6a + 2y = 0 2y - 6a + 2x + 2y = 0 2x + y = 3a x + 2y = 3a ……(2) ….(3)
• 6. Solving Eqs (2) and (3), x = y = a The stationary point is (a, a). Step 2 r = ∂²f/ ∂x² = 4 s = ∂²f/ ∂x ∂y = 2 t = ∂²f/ ∂y² = 4 Step 3 At (a, a), r = 4, s = 2, t = 4 rt - s² = (4)(4) – (2) ² = 12 > 0 Also, r = 4 > 0 Hence, f(x, y) is minimum at (a, a) fmin = a² + a² + (3a - a - a) ² = 3a²
• 7. Example 2 Divide 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 = 120 f = 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 = 40 y = 40 Stationary point is (40, 40).
• 8. Step 2 r = ∂f/ ∂x = -2 s = ∂f/ ∂x ∂y = -1 t = ∂f/ ∂y = -2 Step 3 At (40, 40) rt - s² = (-2)(-2) – (-1) ² = 3 > 0 r = -2 < 0 Hence, f(x, y) is maximum at (40, 40).
• 9. Example 3 Find 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 + z² Since p lines on the surface z² = xy + 1 d² = x² + y² + xy + 1 Let f(x, y) = x² + y² + xy + 1 Step 1 For extreme values , ∂f/ ∂x=0 2x+y=o ∂f/ ∂y=0 2y+x=0 Solving Eqs (1) and (2), x=0 , y=0
• 10. 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 > 0 Also, r = 2 > 0 f(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.
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