Question

Find two positive integers such that their sum is 10, and minimize and maximize the sum of their squares.

Answer 1

Answer:

Minimum: 5, 5

Maximum: 9, 1

Step-by-step explanation:

$x+y=10\\\Rightarrow y=10-x$

The sum of the squares would be

$x^2+y^2=x^2+(10-x)^2\\ =2x^2-20x+100$

$f(x)=2x^2-20x+100$

Differentiating with respect with x

$f'(x)=4x-20$

Equating with zero

$4x-20=0\\\Rightarrow x=\dfrac{20}{4}\\\Rightarrow x=5$

Double derivative of f(x)

$f''(x)=4$

$f''(x)>0$

So, at $x = 5$ the value of the function is minimum.

Therefore, at $x=5$ and $y=10-5=5$ the function has minimum value of $5^2+5^2=50$

For the finding the maximum values of $x$ and $y$ we have to use trial and error method

$9^2+1^2=82$

$8^2+2^2=68$

$7^2+3^2=58$

$6^2+4^2=52$

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.