If we sell \(q\) grams then the revenue is \(4 q\). The break-even point is when revenue is equal to cost, that is, when

\[

4 q=1000+2 q+3 q^{2 / 3}

\]

Let \(f(q)=2 q-3 q^{2 / 3}-1000 .\) We need to solve the equation \(f(q)=0 .\) It is worth asking first whether there is a solution, and whether possibly there might be more than one.

Note that \(f(q)<0\) for "small" values of \(q\), indeed up to 500 and beyond. Also, \(f(1000)>0 .\) Since \(f\) is continuous, it is equal to 0

somewhere between 500 and 1000 . We have \(f^{\prime}(q)=2-q^{-1 / 3}\). From this we conclude easily that \(f\) is increasing from \(q=1 / 8\) on. It follows that \(f(q)=0\) at exactly one place.

The Newton Method yields the recurrence

\[

q_{n+1}=q_{n}-\frac{2 q_{n}-3 q_{n}^{2 / 3}-1000}{2-2 q_{n}^{-1 / 3}}=\frac{q_{n}+1000 q_{n}^{1 / 3}}{2 q_{n}^{1 / 3}-2} .

\]

The simplification is not necessary, but it makes subsequent calculations a bit easier.

Where shall we start? A small amount of experimentation suggests taking \(q_{0}\) to be say 600 . We then get \(q_{1}=607.6089386\), and \(q_{2}=\) \(607.6067886\), perhaps enough to conclude that to the nearest integer the answer is 608. It is, for \(f(607.5)\) and \(f(608.5)\) have different signs.

Note. The problem is slightly easier to handle if we make the substitution \(q=x^{3}\). Then we end up trying to solve the equation \(g(x)=0\) where \(g(x)=2 x^{3}-3 x^{2}-1000\). That way we avoid the unpleasantness of dealing with fractional exponents.