00:01
We have this equation alpha is equal to 2x1 plus 3x2 plus lambda into 10 minus 2x1 square minus 5x2 square.
00:08
So now if we do partial differentiation with respect to x1, do alpha by do x1, we will get 2 minus 4x1 lambda.
00:21
If we equals to this 0, now we can write it as lambda is equal to 2 by 4x1, which is 1 by 2x1 now if we do the partial differentiation with respect to x2 we will get 3 minus 10x2 lambda this is equal to 0 from here lambda will be 3 by 10x2 right since we have two values of lambda we can equate them 1 by 2 x1 is equals to 3 by 10 x2 so this will be, we can write it as x2 is equals to 6x10 x1.
01:12
Right? so now if we do it with respect to lambda, we will get 10 which is equals to 2x1 square plus 5x2 square.
01:31
Now if we put x2 is equals to 6x10 x1 in this equation, we will get 2x1 plus 5 into 6x10 whole square is equal to 10, which is 2x1 plus 5 x6 by 10 whole square is equal to 10, which is 2x1 square plus 5 into 36 by 100 x1 square is equal to 10.
02:03
To 3 .8 x1 square so this will be 3 .8 x1 square is equals to 10 so x 1 square will be 2 .63 so value of x1 is 1 .62 since we have x1 so x2 so x2 into 1 .62 which is 0 .972 right this is the answers for a part now if you say the b part the equation we have this is the equation we have now again if we differentiate with respect to x1 partial differentiation we will get 0 .25 x1 power minus 0 .75 x2 power 0 .75 minus 4x1 lambda is equal to 0 so lambda will be 0 .25 x1 power minus 0 .75 x2 power 0 .75 by 4x1 right so if we differentiate again with respect to x2 we will get 0 .75 x1 power 0 .25 x2 power minus 0 .25 minus 10 lambda x2 is equal to 0 so from here lambda will be 0 .75 x1 power 0 .75 x2 power minus 0 .25 by 10 x2.
04:25
So this is x1 part 0 .1 .25.
04:29
So we have 2 lambda values now again if we equate them.
04:34
0 .25 x1 power.
04:38
Okay, first x2 power 0 .75 into x1 power minus 1.
04:45
0 .75 by 4 x1 is equal to 0 .75 x1 power 0 .25 x1 power 0 .25 x2 power minus 0 .25 by 10x2.
05:07
Right so this will be 0 .25 x2 power 0 .75 by 4 x1 .75 is equal to 0 .75 x1 power 0 .25 by 10 x2 power 1 .25 by 10 x2 power 1 .25 so we can write this as 10 into 0 .25 x2 square is equal to 0 .75 into 4 x1 square.
05:59
This will be x2 square is equal to 1 .2 x1 square.
06:07
So x2 will be x2 will be 1 .09.
06:13
X 1.
06:14
Now if we differentiate with respect to lambda, is equal to 10 is equal to 2 x1 square plus 5 x2 square.
06:28
Again, here if we substitute x2 as 5 into 1.
06:38
X2 x2, this will be 8 x1 square.
06:44
Is equal to 10 so x1 square is 1 .25 value of x1 will be 1 .11 and x2 will be since x2 is 1 .09 x1 into 1 .11 so this is 1 .209 so this is 1 .209 these are the final answer for b part.
07:19
Now if you see the c part, this is the equation we have in the c part...