The output is limited by the item(s) you have least of, hence the $\min$. In general, if producing a unit of output requires $\lambda_i >0 $ of item $i$, then the most output you can create with quantities $x_i$ of item $i$ is $\min_i \frac{1}{\lambda_i} x_i$.
To answer the original question, if item $i$ costs $c_i>0$ and the total available capital is $C>0$, then the maximum you can produce is $\frac{C}{\sum_i c_i \lambda_i}$.
The problem is a standard linear programming problem, in this case the solution can be obtained by inspection. However, it is fairly easy to see why this is a solution:
The problem is $\max\{ \min_i \frac{1}{\lambda_i} x_i | \sum_i c_i x_i \leq C, x_i \geq 0 \}$. Since the $c_i >0$, the solution space is compact, and the objective is continuous, hence a solution exists. Life becomes a little easier if we rescale the problem with $y_i = \frac{1}{\lambda_i} x_i$, then the problem is to solve $\max\{ \min_i y_i | \sum_i c_i \lambda_i y_i \leq C, y_i \geq 0 \}$. Since $y=0$ is feasible, we can see that this is the same problem as $\max\{ \min_i y_i | \sum_i c_i \lambda_i y_i \leq C\}$. If $\sum_i c_i \lambda_i y_i < C$, then it is clear that each $y_k$ can be increased slightly to increase the objective, hence the problem is equivalent to $\max\{ \min_i y_i | \sum_i c_i \lambda_i y_i = C\}$.
Suppose for some index $j_0$ we have $y_{j_0} > \min_i y_i$. Let $K = \{k| y_k = \min_i y_i\}$. Let $h$ be the vector $h_k = 1$ for $k\in K$, $h_{j_0} = -\frac{\sum_{k \in K} \lambda_k c_k}{\lambda_{j_0}c_{j_0}}$, and $h_j = 0$ otherwise. Note that $\sum_i c_i \lambda_i (y_i+t h_i) = C$, and for sufficiently small positive $t>0$, $\min_i y_i < \min_i (y_i+t h_i)$. Hence at a solution to the problem, we have $y_i = \hat{y}$ for all $i$. Then the constraint gives $\sum_i c_i \lambda_i \hat{y} = C$, from which we obtain the solution $\max\{ \min_i \frac{1}{\lambda_i} x_i | \sum_i c_i x_i \leq C, x_i \geq 0 \}= \frac{C}{\sum_i c_i \lambda_i}$.
Best Answer
Hint:
$Q$ is a function of $p$, and $p$ is a function of $t$. What do you get when you insert the expression $p=0.04t^2+0.2t+12$ into $Q$?