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The difference of two positive numbers is 9. What is the minimum sum of their squares?
Absolute extrema/optimization
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Let x be a greater of these two real numbers; y be a smaller of these two real numbers.
Then x - y = 9, or x = y+9.
They want you find the minimum of , which is
= = .
So, actually, we want to find the minimum of this quadratic function
f(y) = on the set of positive real numbers y > 0.
The global minimum of a quadratic function of the general form f(y) = ay^2 + by + c with positive
leading coefficient "a" is achieved at = . In this case
= = -4.5,
which is negative value out of the scope of our consideration.
In the area y > 0, we have the ascending branch of the parabola on the right from its minimum in negative
number. Had we consider the scope of non-negative real numbers y >= 0, then the minimum of the parabola
f(y) = would be at y = 0 with the value f(0) = 81.
But, according to the problem, our scope is the set of positive numbers y > 0, and in this domain,
quadratic function f(y) = does not have the minimum, at all.
ANSWER. In the set of positive numbers, the function with the restriction x-y = 9 has no minimum.
Solved.
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Above, I told you the whole story with all details.
I could explain everything in more short terms, simply noticing that
the parabola f(y) = 2y^2 + 18y + 81 has two positive terms 2y^2 and 18y
in the domain y > 0.
So, the values of f(y) go up as "y" goes to the right from zero,
which means that function f(y) raises there.
In opposite, function f(y) goes down as "y" approaches 0 from the right side.
But "y" never gets 0 in the positive domain - so we never get the minimum of f(y)
in the positive domain y > 0.
With these two explanations, you have now a complete picture and a complete vision to the problem.