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

CC BY-SA 4.0
2022-06-19T16:39:48.313Z
2022-06-19T16:45:48.360Z
2022-06-19T16:45:48.360Z
Kur_Kush
<p>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
<span class="math-container">$$
C = 0.1{q}^{2}+q+20
$$</span>
where <span class="math-container">$q$</span> 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?
<span class="math-container">$$ --------------------------------- $$</span></p>
<p>I think we need an &quot;expected revenue&quot; function, but this is what I'm unsure about because my textbook has no examples like this. I'm assuming it's
<span class="math-container">$$
R = 7*{q}_{cold} + 5*{q}_{warm} \\ R = 7*0.55*q + 5*0.45*q \\ R = 6.1q
$$</span>
Then we maximize profit on quantity:
<span class="math-container">$$\pi = R-C \\ \pi = 6.1q-(0.1{q}^{2}+q+20) 
\\ \pi = - 0.1{q}^{2}+5.1q -20 
\\ \frac{d\pi}{dq} = -0.2q + 5.1 = 0
\\ q = 25.5
$$</span>
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.</p>

CC BY-SA 4.0
2021-11-13T23:32:24.720Z
Try maximize expected *profit* instead of expected revenue.
Herr K.
CC BY-SA 4.0
2021-11-14T02:09:31.637Z
@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.
kleinerde
CC BY-SA 4.0
2021-11-14T08:25:35.217Z
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
CC BY-SA 4.0
2021-11-14T16:40:30.873Z
@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.
kleinerde
<profit-maximization><probability><cost-functions>
Maximizing profit with a simple probabilistic production function (basic practice problem)
kleinerde
# Maximizing profit with a simple probabilistic production function (basic practice problem)

(Asked by: username_0 on 2021-11-13)

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.

**Tags:** profit-maximization, probability, cost-functions

> **Comments:**
> **username_1**: Try maximize expected *profit* instead of expected revenue.  
> **username_0**: @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.  
> **username_2**: 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.  
> **username_0**: @username_2 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.  

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## 1 answers

### Answer by username_3 on 2022-06-19. Score: 1

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 $.


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