# Maximizing profit with a simple probabilistic production function (basic practice problem)

A restaurant finds that less orders for their soup of the day are placed on warmer days so they discount the usual 7USD price to 5USD on warmer days. The cost of making the soup is given by $$C = 0.1{q}^{2}+q+20$$ where $$q$$ is the bowls of soup per day and the probability of a warm day is 0.45. How much soup should the restaurant make if it wants to maximize profits from the soup of the day? $$---------------------------------$$

I think we need an "expected revenue" function, but this is what I'm unsure about because my textbook has no examples like this. I'm assuming it's $$R = 7*{q}_{cold} + 5*{q}_{warm} \\ R = 7*0.55*q + 5*0.45*q \\ R = 6.1q$$ Then we maximize profit on quantity: $$\pi = R-C \\ \pi = 6.1q-(0.1{q}^{2}+q+20) \\ \pi = - 0.1{q}^{2}+5.1q -20 \\ \frac{d\pi}{dq} = -0.2q + 5.1 = 0 \\ q = 25.5$$ So the restaurant should usually prepare soup for 25.5 bowls, given the uncertainty about the weather. I have no answer keys, so I'm looking for a confirmation or correction.

• Try maximize expected profit instead of expected revenue. Nov 13, 2021 at 23:32
• @HerrK. One might regret doing it- but here I don't know if it works. I think we need a budget constraint. I get a maximized ratio of 9 to 11 for warm day bowls to cold day bowls. Nov 14, 2021 at 2:09
• Does the restraurant know the weather before making the soup or not? If the first is the case, the amount made should be conditional on the weather. If not $q$ should be independent of the weather and you shoul indeed maximise expected profits.
– tdm
Nov 14, 2021 at 8:25
• @tdm In reality, for this particular problem, we would have a good weather forecast for the next day and a generally good forecast for the week. The warm day probability will change over the year, so if you maximized by the week you would need to make this probability variable (ok - to do). Otherwise, over the year, the maximized ratio above (if correct) would hold. Nov 14, 2021 at 16:40

For a large $$n$$, the quantity sold in $$n$$ days will be $$0.45n * 20 + 0.55n * 30 = 25.5n$$ considering there'll be $$0.45n$$ warm days where the price/cup will be $$\\\ 5$$ and there'll be $$0.55n$$ cold days where the price/cup will be $$\\\ 7$$. On an average, you'll be selling $$\frac{25.5n}{n} = 25.5$$ cups/day.
This means the quantity sold per day, on an average, will be $$25.5$$ and the profit (per day) will be $$\\\ \frac{0.45n * 20 + 0.55n * 70}{n} = \\\ \frac{47.5n}{n} =\\\ 47.5$$.