By simply expanding to get a quadratic expression in and, it can be shown that the values of and that minimize the objective function are
2.
Since this is a quadratic expression, the vector which gives the global minimum may be found via matrix calculus by differentiating with respect to the vector " b " ( using denominator layout ) and setting equal to zero:
3.
Given a value of A and ignoring the contributions of nucleon spin pairing ( i . e . ignoring the \ pm \ delta ( A, Z ) term ), the binding energy is a quadratic expression in Z that is minimized when the neutron-proton ratio is N / Z \ approx 1 + \ frac { a _ C } { 2a _ A } A ^ { 2 / 3 }.
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