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Solve for reals: $$a^2+b^2-ab-b+\frac 13=0$$

My attempt.

enter image description here

$$\frac {d}{da} \left(a^2+b^2-ab-b+\frac 13\right)=0$$

$$2a-b=0$$

$$\frac {d}{db} \left(a^2+b^2-ab-b+\frac 13\right)=0$$

$$2b-a-1=0$$

And I got $a=\frac 13, b=\frac 23$

If I define $$f(a,b)=a^2+b^2-ab-b+\frac 13$$

Then is it correct using partial derivatives to find $(a,b)$ ?

I am aware completing the square method. But, I want to know the errors of this method I use.

Here are a graph $z = x^2 - xy + y^2 - y$ in 3-D (vertical is distorted, not sure how to fix that, although carefully adjusting $x,y$ bounds might work)

Also a contour plot. Ellipses

Found out how to adjust the $x,y$ bounds, this version seems better enter image description here

enter image description here

Will Jagy
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user1094359
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2 Answers2

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No, it's wrong. I think the problem is within a naive understanding of the terms you're using. Think about it this way and I hope it will help you:

You attempted to differentiate by $a$, and by some induction I suppose we can agree that it is as valid as differentiating by $b$. Then ask, what are the objects that we can differentiate at all? They are functions! Not the expressions. What are the symbols we use in the differentiation? They are the arguments (I'll call them parameters) of the function. So since you can't differentiate a zero, but only a function that is zero everywhere and since we are assuming that the functions are parametrized (I am being a little bit informal here) by $a$ and $b$, we deduce that the zero represents a function (in terms) of $a$ and $b$ that is equal to zero everywhere. Then the implication $f=g\Rightarrow\partial f=\partial g$ (one by which you allowed yourself to deduce $\partial f=\partial g$) means that we are assuming that the functions are equal. But in your case they are clearly not equal because it is easy to find a pair of $a$ and $b$ (formally, an element of the domain) for which they differ: for example $a=1$ and $b=0$.

donaastor
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$$a^2+b^2-ab-b+ \frac{1}{3} = \left( a - \frac{b}{2} \right)^2 + \frac{3}{4} \left( b - \frac{2}{3} \right)^2 $$

Will Jagy
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    Please, see my question. I am aware completing the square, I mentioned this in my question. – user1094359 Nov 28 '22 at 21:19
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    @user1094359 I stopped reading once I saw that you were differentiating something. – Will Jagy Nov 28 '22 at 21:34
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    OP explicitly wrote “I am aware completing the square method. But, I want to know the errors of this method I use.” So this does not answer the question at all. – Martin R Nov 29 '22 at 08:40