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Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
Differentiating equation . , we get the gradient to be: 2xi + 2yj
Substituting (x = 1) and (y = 2 2), we get the gradient to be: 2i + 4 2 j
2i + 4 2 j 2i + 4 2 j i 2 2 j
The direction of the gradient is given by: = = + - - - - (1.80)
4 + 16 ∗ 2 6 3 3
It can be observed that equation (1.80) is the direction of the normal.
Therefore, the above property of the gradient is used in optimization to find minimum or maximum
value of a function by continuous search along the local gradient directions.
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Example: locating the hottest point in a block which must lie on a circle of certain radius.
In general, if a function (f(x,y)) is to be optimized with a constraint [ax + by =c], that the
Lagrange (L) is given by:
L = f x, y + λ ax + by − c - - - - (1.81)
The above equation is solved by taking the partial derivative of (L) with respect to (x, y and λ),
where (λ) is the Lagrangian multiplier. The values of (x, y and λ) are got by simultaneously
solving the partial derivatives of (x, y and λ), as follows:
∂L
= 0 - - - - (1.82)
∂x
∂L
= 0 - - - - (1.83)
∂y
∂L
= 0 - - - - (1.84)
∂λ
Simultaneously solving equation (1.82), (1.83) and (1.84) we get the values of (x, y and λ),
which on substitution with equation (1.81), would give the optimum value of the function.
Example 1: consider the perimeter of a rectangle to be 12m. Find the optimum area of the
rectangle using Lagrangian multipliers.
The function to be optimized is the area of the rectangle. Let ‘x’ and ‘y’ be the sides of the
rectangle and the perimeter is given by (2x + 2y = 12).
The Lagrangian L is given by: xy + λ 2x + 2y − 12 - - - - (1.85)
QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 63
