I know for a fact that this relation is $x > x^2$ but I am having trouble interpreting "by the minimum amount". is it minima? if so how do I express it as a function? in order to find the answer using calculus?

I"m assuming you mean maximum. If you mean minimum, you may let $x=1$ where $x^2=x$.

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You want to maximize $f(x)=x-x^2$. You can differentiate, set to $0$ and you get $ \frac12$. You need to see if this is a maximum, which it is.

In other words, we need to minimise $$x-x^2$$ or $$\dfrac {1}{4}-\left (x-\dfrac {1}{2}\right )^2$$ which you can do by maximising the square, or in other words, making $x\to \infty$. However if you look to maximise the difference, you need to minimise the square, or in other words, make the square term zero, as the least value a square can take is $0$. This happens when $x=\dfrac 12$.

Firstly it should be by the maximum amount , since you have given us the answer $\frac{1}{2}$ . For that just derive the function $x-x^2$ and set the derivative $= 0 $. Since the second derivative is -ve , therefore it is maximum at $\frac{1}{2}$

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Divide a number N into two parts in such a way that three times the square of one part plus twice the square of the other part shall be a minimum.

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