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Given price, marginal cost and fixed cost, find the quantity that maximises profits.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "A demand function for a generic good, where $P$ is the price and $q$ is the quantity is: $P = \\var{p} - q$. The Marginal Cost is given by, $MC = \\simplify{{x}q+{y}}$, and fixed cost is, $FC = \\var{z}$. Find the quantity $q$ to maximise profits.
", "advice": "The total profit can be found by calculating the total revenue (price times quantity) subtract the total cost (marginal cost times quantity add fixed costs).
\nAs an equation this looks like:
\n\\[ \\begin{split} \\pi &= TR - TC \\\\ &= Pq - (MC \\times q + FC) \\end{split} \\]
\nUsing the equations in this question we have:
\n\\[ \\pi = \\simplify{({p} -q)q - ({x}q+{y})q - {z}} \\]
\nCollecting like terms:
\n\\[ \\pi = \\simplify[!noLeadingMinus]{-{x+1}q^2 + {p-y}q +{z}} \\]
\nDifferentiating with respect to $q$ gives:
\n\\[ \\frac{d\\pi}{dq} = \\var{-2*(x+1)}q + \\var{p-y} \\]
\nUsing the first order condition $ \\frac{d\\pi}{dq} = 0 $ we have:
\n\\[ \\var{-2*(x+1)}q + \\var{p-y}= 0 \\]
\nSolving to find $q$:
\n\\[ q = \\var{-(p-y)/(-2*(x+1))} \\]
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