Answer :
x = 8 and y = 8 are the two positive numbers whose sum is 16 and the sum of the squares is minimum.
According to the question,
The two numbers are positive and the sum of the two numbers is 16 and the sum of the squares of the two numbers is the minimum.
Now to solve the question we will 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
x + y = 16 y = (16 - x)
Assume, S = x² + y²
S = x² + (16 - x)²
Now we will differentiate with respect to x
dS/dx = 2x+ 2(16-x)(-1)
2x-2(16-x)=0
2x-32+2x=0
4x-32=0
4x=32
x=32/4
x=8
As x > 0, x = 8
d²S/dx² = d/dx [2x+ 2(16-x)(-1)]
d²S/dx² = d/dx (2x) - 2 d/dx (16-x)
d²S/dx² = 2 - 2[0-1]
d²S/dx² = 4
Now when x=8
d²S/dx² = 4 ≥0
Therefore, the function S = sum of the squares of the two numbers is minimum at x = 8.
y = 16-8 = 8
Therefore, x = 8 and y = 8 are the two positive numbers whose sum is 16 and the sum of the squares is minimum.
To learn more about maximum and minimum function and differentiation,
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