Find two positive numbers whose sum is 16 and the sum of whose squares is minimum.

Given,


The two numbers are positive.


the sum of two numbers is 16.


the sum of the squares of two numbers is minimum.


Let us consider,


x and y are the two numbers, such that x > 0 and y > 0


Sum of the numbers : x + y = 16


Sum of squares of the numbers : S = x2 + y2


Now as,


x + y = 16


y = (16-x) ------ (1)


Consider,


S = x2 + y2


By substituting (1), we have


S = x2 + (16-y)2 ------ (2)


For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function f(x) has a maximum/minimum at a point c then f’(c) = 0.


Differentiating the equation (2) with x




----- (3)


[Since ]


Now equating the first derivative to zero will give the critical point c.


So,



2x – 2(16 -x) = 0


2x – 32 + 2x = 0


= 4x = 32



x = 8


As x > 0, x = 8


Now, for checking if the value of S is maximum or minimum at x=8, we will perform the second differentiation and check the value of at the critical value x = 8.


Performing the second differentiation on the equation (3) with respect to x.





[Since and ]



Now when x = 8,



As second differential is positive, hence the critical point x = 8 will be the minimum point of the function S.


Therefore, the function S = sum of the squares of the two numbers is minimum at x = 8.


From Equation (1), if x= 8


y = 16 – 8 = 8


Therefore, x = 8 and y = 8 are the two positive numbers whose su is 16 and the sum of the squares is minimum.


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