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It costs a firm $C(q)$ dollars to produce $q$ grams per day of a certain chemical, where
$C(q)=1000+2 q+3 q^{2 / 3}$
The firm can sell any amount of the chemical at $\ 4$ a gram. Find the break-even point of the firm, that is, how much it should produce per day in order to have neither a profit nor a loss. Use the Newton Method and give the answer to the nearest gram.
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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.
by Platinum (130,878 points)

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