 ह jailay evriven compilerat meiself ASMR assistant professor Department of Humanities & Sciences Vaalchan Institute of Technology Sa片wpper Y a k i t ne lagranj won ensuite entangel sam 98ames stamina ki Mad각 Learning outcome At the end of this session students will be able to find these stationary values of given function subjected to some conditions. Now actually the problem is we have given one function F which may be a function of 2 or 3 independent variable and with that function we have provided one restriction. Now in this problem we want to extremize the given function subjected to certain condition to find the extreme value in such a problem we will use lagrange's undetermined multipliers now in the step number 1 first we will denote the given function by F here we have considered this F as a function of 3 independent variables call it as equation number 1 and we will denote the restriction or the subjected condition by the notation 5 here also we have considered it as a function of x, y, z and every time we will write in this form 5 of x, y, z equal to 0 and call it as equation number 2 in the step number 2 we have to consider these 3 equations which are called as lagrange's equations now call them as equation number 3 now the step number 3 is we have to solve these 2 equations for the values of x, y, z by eliminating lambda once we eliminate the lambda and get the values of x, y, z that values of x, y, z are called the stationary values of this given function 1 and if we substitute these values in this equation 1 we will get the extreme that is maximum or minimum value of this function subjected to this condition now pause this video and find the stationary value of a function x, y subjected to the condition x plus y equal to 16 was lagrange's equation r x plus lambda equal to 0 and y plus lambda equal to 0 I hope that all of you have written the answer the problem is to finding the stationary values where the lagrange's equations are provided here now the first one is x plus lambda equal to 0 from this equation directly we get the value of x as minus lambda and second lagrange's equation is y plus lambda equal to 0 which implies y equal to minus lambda now to eliminate this lambda we will use this condition x plus y equal to 16 now substituting the values of x and y in this equation we get x as minus lambda y as also minus lambda equal to 16 which implies minus 2 lambda equal to 16 it implies after dividing by minus 2 lambda equal to minus 8 now replacing the value of lambda in these 2 equations we get the value of x and y as 8 and 8 now which is the required stationary value of the given function x into y let us consider the first example find the maximum and minimum values of the function f of x, y is equal to 3x plus 4y on the circle x square plus y square equal to 1 using the method of Lagrange's multipliers solution now in the step number 1 now it is the given function which is to be extremized therefore we will denote it by small f call it as equation number 1 this is the restriction we will write it in a standard form by taking this one to the right hand side and we will denote it by phi as x square plus y square minus 1 is equal to 0 now call it as equation number 2 now here the f and the restriction are in terms of 2 independent variables therefore we have 2 Lagrange's equations now that Lagrange's equations are the first one is doh f by doh x plus lambda times doh phi by doh x equal to 0 it implies now differentiating this f partially with respect to x we get it as 3 plus now differentiating this phi partially with respect to x we get it as 2x but already lambda is there therefore it is 2x lambda and equal to 0 now let us call it as equation number 3 now one more Lagrange's equation we have doh f by doh y plus lambda times doh phi by doh y equal to 0 which implies now differentiating this f partially with respect to y treating remaining variables constant we get the value as 4 plus lambda times now differentiating phi partially with respect to y its derivative is 0 and y square has the derivative 2y and this lambda as it is and minus 1 has the derivative 0 therefore it is equal to 0 now let us call it as equation number 4 now in the next step our aim is to eliminate lambda from this 2, 3 and 4 and get the values of x, y, z which are our required stationary values now to do so from 3 we can see that it is possible to obtain the value of x and which is minus 3 by 2 lambda call it as equation number 5 now from this 4 it is possible to find the value of y as minus 2 by lambda call it as equation number 6 in order to eliminate this lambda we will use this condition number 2 we will substitute the value of x and y here now putting the value of x and y into we get it is x square and the value of x is minus 3 by 2 lambda square plus next term is y square and y is minus 2 by lambda bracket square and minus 1 as it is equal to 0 now the square of this term is 9 upon 4 lambda square plus and its square is 4 upon lambda square minus 1 is equal to 0 we can make the same denominator here by multiplying the denominator by 4 to the second term we get here 4 into 4 is 16 and here it is 4 lambda square now 16 plus 9 is 25 upon 4 lambda square minus 1 is equal to 0 now it implies 25 upon 4 lambda square is equal to 1 after taking 1 to the right hand side now taking lambda square to the right hand side we get lambda square is equal to 25 upon 4 but we want the value of lambda it implies lambda is equal to 5 by 2 minus 5 by 2 thus we have obtained 2 values of lambda one is 5 by 2 another is minus 5 by 2 we will substitute the values of lambda in 5 and 6 to get the stationary values as we have obtained 2 values we will consider them in 2 different cases now the case number 1 consider the first value of lambda as 5 by 2 now from 5 replacing lambda by 5 by 2 we get x equal to minus 3 by 2 into lambda, lambda is 5 by 2 which is equal to we can remove these 2 we get the value as minus 3 by 5 now from 6 putting the value of lambda y equal to minus 2 by it is lambda, lambda is 5 by 2 which is equal to this 2 is in the denominator of denominator we can write it in numerator so it is minus 4 by the first pair of x y which we have obtained as minus 3 by 5 and minus 4 by 5 now this is the stationary value to find the extreme value let us substitute the stationary value in the given function 1 therefore at the stationary point minus 3 by 5 minus 4 by 5 which has the value 3 into x is minus 3 by 5 plus 4 into y is minus 4 by 5 which is equal to 3 into minus 3 is minus 9 by 5 and it is 4 into minus 4 is minus 16 by 5 that is equal to denominator is same we can add the numerators so minus 16 minus 9 minus 25 by 5 which is equal to minus 5 the value of the given function which is obtained at this stationary point we have one more value of lambda we will consider it in case number 2 consider lambda equal to minus 5 by 2 again from 5 we get the value of x equal to minus 3 by 2 into minus 5 by 2 2 will be removed and minus minus plus 3 by 5 from 6 replacing the value of lambda is minus 5 by 2 minus 5 by 2 2 upon in bracket minus 5 by 2 minus minus will be plus now this 2 will be in the numerator so it is plus 4 by 5 now this is 3 by 5 and 4 by 5 is the next stationary value let us calculate the value of the function at this stationary point that is add 3 by 5, 4 by 5 f is 3x plus 4y substituting the values we get 3 by 5 plus 4 into 4 by 5 which is equal to 9 by 5 plus 16 by 5 as the denominators are same adding the numerators we get 16 plus 9 is 25 by 5 which is equal to 5 thus we have obtained 2 extreme values of given function one is minus 5 and another one is plus 5 so obviously we can say that the minimum value subjected to the given condition is minus 5 and the maximum value is 5