Thread: Differential calculus of higher variables

if x1.x2.x3.....xn=1 where xi is positive then show that x1+x2+x3+.....+xn >=n (greater than or equal to)

Guys need your help

This was given as a problem in the chapter of Differential calculus of functions of several variables. How to solve it

Originally Posted by vinmech32

if x1.x2.x3.....xn=1 where xi is positive then show that x1+x2+x3+.....+xn >=n (greater than or equal to)

Guys need your help

This was given as a problem in the chapter of Differential calculus of functions of several variables. How to solve it

This result is an easy application of the arithmetic-geometric mean inequality. But if you want to use it as an exercise in several-variable calculus, then you can regard it as asking you to find the minimum value of the function $\displaystyle f(x_1,x_2,\ldots,x_n) = x_1+x_2+\ldots +x_n$ in the region $\displaystyle \{x_i>0: 1\leqslant i\leqslant n\}$, subject to the constraint $\displaystyle x_1x_2\cdots x_n=1.$ It then becomes an exercise in the use of Lagrange multipliers, and you can find that the minimum value of f is n, occurring at the point $\displaystyle (x_1,x_2,\ldots,x_n) = (1,1,\ldots,1).$

Originally Posted by Opalg

It then becomes an exercise in the use of Lagrange multipliers

... unless you wrote $\displaystyle x_n$ as $\displaystyle \frac{1}{x_1 x_2 \ldots x_{n-1}}$,

and minimised

$\displaystyle y\ =\ x_1 + x_2 + \ldots + x_{n-1} + \frac{1}{x_1 x_2 \ldots x_{n-1}}$

Setting all the first derivatives equal to zero makes all the x's equal to one, and then you can show the Hessian is positive definite at that point.

Originally Posted by Opalg

This result is an easy application of the arithmetic-geometric mean inequality. But if you want to use it as an exercise in several-variable calculus, then you can regard it as asking you to find the minimum value of the function $\displaystyle f(x_1,x_2,\ldots,x_n) = x_1+x_2+\ldots +x_n$ in the region $\displaystyle \{x_i>0: 1\leqslant i\leqslant n\}$, subject to the constraint $\displaystyle x_1x_2\cdots x_n=1.$ It then becomes an exercise in the use of Lagrange multipliers, and you can find that the minimum value of f is n, occurring at the point $\displaystyle (x_1,x_2,\ldots,x_n) = (1,1,\ldots,1).$

Thanks for the reply.. Can u pls brief me the process of evaluating it by the use of lagrange multipliers...(any examples)

Originally Posted by vinmech32

Can u pls brief me the process of evaluating it by the use of lagrange multipliers...(any examples)

Let
$\displaystyle \begin{aligned}F(x_1,x_2,\ldots,x_n,\lambda) &= f(x_1,x_2,\ldots,x_n) + \lambda(x_1x_2\cdots x_n-1) \\ &= x_1+x_2+\ldots +x_n + \lambda(x_1x_2\cdots x_n-1).\end{aligned}$

Put the partial derivative with respect to $\displaystyle x_i$ equal to 0:

$\displaystyle \frac{\partial F}{\partial x_i} = 1 + \lambda x_1x_2\cdots x_{i-1}x_{i+1}\cdots x_n = 0.$

Multiply by $\displaystyle x_i$ to see that $\displaystyle x_i = -\lambda$ (because $\displaystyle x_1x_2\cdots x_n=1$). Thus all the $\displaystyle x_i$'s are equal, and since their product is 1 they must all be equal to 1. Therefore the only critical point is $\displaystyle (x_1,x_2,\ldots,x_n) = (1,1,\ldots,1).$

You can find other worked examples here.

thanks guys for your assistance.....
u r doing a great job...