Two numbers whose sum is 10 is required.
The answer is
Minimum: 5, 5
Maximum: 9, 1
Let [tex]x[/tex] and [tex]y[/tex] be two numbers
[tex]x+y=10\\\Rightarrow y=10-x[/tex]
The sum of the squares would be
[tex]x^2+y^2=x^2+(10-x)^2\\ =2x^2-20x+100[/tex]
[tex]f(x)=2x^2-20x+100[/tex]
Differentiating with respect with x
[tex]f'(x)=4x-20[/tex]
Equating with zero
[tex]4x-20=0\\\Rightarrow x=\dfrac{20}{4}\\\Rightarrow x=5[/tex]
Double derivative of f(x)
[tex]f''(x)=4[/tex]
[tex]f''(x)>0[/tex]
So, at [tex]x = 5[/tex] the value of the function is minimum.
Therefore, at [tex]x=5[/tex] and [tex]y=10-5=5[/tex] the function has minimum value of [tex]5^2+5^2=50[/tex]
For the finding the maximum values of [tex]x[/tex] and [tex]y[/tex] we have to use trial and error method
[tex]9^2+1^2=82[/tex]
[tex]8^2+2^2=68[/tex]
[tex]7^2+3^2=58[/tex]
[tex]6^2+4^2=52[/tex]
0 and 10 will not be considered as 0 is neither positive nor negative.
So, the maximum value of the numbers is 9 and 1 and the maximum value of the function is 82.
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