$$ C(q)=0.01 q^{3}-0.6 q^{2}+14 q $$ What is the fixed cost? fixed cost $$ =0 $$ What is the maximum profit if each item is sold for 10 dollars? (Assume you sell everything you produce.) maximum profit $$ =167.04 $$ Suppose we fix production at 36 goods produced, and that they all sell when the price is 10 dollars each. Also suppose that for each 1 dollar increase in price, 2 fewer goods are sold (so if the price is 11 dollars, 34 of the 36 goods being produced are sold.) To maximize profit in this case, we should increase (Enter zero if the price should remain at 10 dollars.)

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We first note that when $$ q=36 $$ goods are produced, the total cost is determined solely by production. The cost function given is  
$$
C(q)=0.01q^3-0.6q^2+14q.
$$

Since we fix production at 36 goods, the cost is  
$$
C(36)=0.01(36)^3-0.6 (36)^2+14(36).
$$
Calculate each term:
- $$ (36)^3=46656 $$ so $$ 0.01(36)^3=466.56 $$.
- $$ (36)^2=1296 $$ so $$ 0.6(36)^2=777.6 $$.
- $$ 14\times 36=504 $$.

Thus,  
$$
C(36)=466.56-777.6+504=192.96.
$$

When the price is 10 dollars and all 36 goods sell, the revenue is  
$$
R=10\times36=360.
$$
So the profit at 10 dollars per item is  
$$
\pi=360-192.96=167.04.
$$

Now we consider the scenario where raising the price causes fewer goods to sell. Suppose we increase the price by $$ x $$ dollars above 10. Then:  
- The new price is $$ 10+x $$.
- For each 1 dollar increase, 2 fewer goods are sold, so the number sold is  
$$
36-2x.
$$

The revenue as a function of $$ x $$ is  
$$
R(x) = (10+x)(36-2x).
$$
Expanding this, we get  
$$
R(x)=10(36-2x)+x(36-2x)=360-20x+36x-2x^2=360+16x-2x^2.
$$

Since the cost of production remains $$ 192.96 $$ for 36 goods, the profit function is  
$$
\pi(x)=R(x)-192.96=360+16x-2x^2-192.96.
$$
Simplify the constant terms:  
$$
360-192.96=167.04,
$$
so  
$$
\pi(x)=167.04+16x-2x^2.
$$

This is a quadratic function in $$ x $$ with a negative coefficient for $$ x^2 $$, indicating a concave down parabola that achieves a maximum at its vertex.

The vertex of a quadratic function $$ \pi(x)=ax^2+bx+c $$ occurs at  
$$
x=-\frac{b}{2a}.
$$
Here, $$ a=-2 $$ and $$ b=16 $$. Thus,  
$$
x=-\frac{16}{2(-2)}=-\frac{16}{-4}=4.
$$

A value of $$ x=4 $$ means we should increase the price by 4 dollars above 10 dollars, so the new price is $$ 10+4=14 $$ dollars.

Hence, to maximize profit in this case, we should increase the price by  
$$
\boxed{4}.
$$
To maximize profit when producing 36 goods at 10 dollars each, increase the price by 4 dollars to 14 dollars.

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