## Question

Using Lagrange multipliers, show that:

a) maximum value of subject to is

b) minimum value of subject to is

## Solution

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### a) maximum value of subject to is

Objective function,

Constraint function,

Lagrange multiplier function:

__Step-1: Find Gradient of f__

__Step-2: Find Gradient of g__

__Step-3: Find x, y and __

__Step-4: Substitute and figure out function values__

Let’s take as the question is to prove maxima.

Let c = ,

Apply c in Constraint function

Apply (7) in (1)

*Thus, f has maxima at . Hence, proved.*

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### b) minimum value of subject to is

Objective function,

Constraint function,

Lagrange multiplier function:

__Step-1: Find Gradient of f__

__Step-2: Find Gradient of g__

__Step-3: Find x, y and __

(1)

Similarly, we get,

(2)

(3)

RHS of the equations (1), (2) and (3) are the same. Thus, let’s equate LHS of (1) = (2)

Similarly, (2)=(3) and (1)=(3) will give =1/x, =1/z

Thus, x=y=z=

__Step-4: Substitute and figure out function values__

Apply in Constraint function

Let

Let’s take as the question is about a minima.

Apply in Objective function

*Thus, f has minima at . Hence, proved.*