Q. 64 The arithmetic mean of the real ... [FREE SOLUTION]

Chapter 12: Q. 64 (page 987)

The arithmetic mean of the real numbers $a_{1}, a_{2}, . . . ., a_{n}$ is $\frac{1}{n} (a_{1} + a_{2} + . . . . + a_{n})$ . If $a_{i} > 0$ for 1 鈮� i 鈮� n, then the geometric mean of $a_{1}, a_{2}, . . . ., a_{n} i s {(a_{1} a_{2} . . . a_{n})}^{\frac{1}{n}}$ . In Exercises 64鈥�66 we ask you to prove that the geometric mean is always less than the arithmetic mean for a set of positive numbers.
Use the method of Lagrange multipliers to show that $\sqrt{x y} 猢� \frac{1}{2} (x + y)$ when x and y are both positive.

Short Answer

Expert verified

$A = \frac{x + y}{2}, f (x, y) = \sqrt{x y}, g (x, y) = \frac{x + y}{2} - A . B y t h e m e t h o d o f L a g r a n g e m u l t i p l i e r : 鈭� f = 位鈭� g . S o l v i n g a n d e q u a t i n g t h e v a l u e s o f 位 w e g e t, x = y . P u t t i n g i n e q u a t i o n A = \frac{x + y}{2}, w e g e t, x = y = A . H e n c e, \sqrt{x y} 猢� \frac{x + y}{2} .$

Step by step solution

Step 1. Given Information.

Given: AM of two numbers a and b is: $\frac{a + b}{2} .$

And GM of two numbers a and b is: $\sqrt{a b}$ .

Step 2. Proof by method of Lagrange's method.

$L e t A = \frac{x + y}{2} . T h e f u n c t i o n s : f (x, y) = \sqrt{x y} a n d g (x, y) = \frac{x + y}{2} . T h e c o n s t r a i n t f u n c t i o n : g (x, y) = \frac{x + y}{2} - A . b y t h e m e t h o d o f L a g r a n g e m u l t i p l i e r s, 鈭� f = 位鈭� g, w e g e t, <\frac{鈭� f}{鈭� x}, \frac{鈭� f}{鈭� y}> = 位 <\frac{鈭� g}{鈭� x}, \frac{鈭� g}{鈭� y}> <\frac{\sqrt{y}}{2 \sqrt{x}}, \frac{\sqrt{x}}{2 \sqrt{y}}> = <\frac{位}{2}, \frac{位}{2}> E q u a t i n g t h e c o r r e s p o n d i n g v a l u e s, \frac{\sqrt{y}}{2 \sqrt{x}} = \frac{位}{2} a n d \frac{\sqrt{x}}{2 \sqrt{y}} = \frac{位}{2} N o w e q u a t i n g t h e v a l u e o f 位, \frac{\sqrt{x}}{2 \sqrt{y}} = \frac{\sqrt{y}}{2 \sqrt{x}} W e g e t, x = y . S u b s i t u t i n g t h e v a l u e o f x o r y i n A = \frac{x + y}{2}, w e g e t, x = y = A . T h u s, t h e m a x i m u m v a l u e o f f (x, y) i s a t A . H e n c e, \sqrt{x y} 猢� \frac{x + y}{2} .$

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Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Proof by method of Lagrange's method.

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