Faced with a 3000 taco capacity constraint, what is the optimal number of chicken tacos to prepare?

Steps to Solve

1. Define Variables

Let $x$ be the number of chicken tacos prepared.

2. Define the Demand Distribution

The demand for chicken tacos follows a uniform distribution between 60 and 160.

3. Define the Cost Parameters

• Cost per chicken taco = $0.75
• Selling price per chicken taco = $1.75
• Salvage value per unsold chicken taco = $0.50

4. Setup the Profit Function

The profit $P(x)$ can be defined based on whether demand exceeds the prepared quantity or not. Let $D$ be the demand.

$$ P(x) = \begin{cases}

1. 75x - 0.75x, & \text{if } D \geq x \text{ (All tacos are sold)} \

2. 75D + 0.50(x-D) - 0.75x, & \text{if } D < x \text{ (Some tacos are unsold)} \end{cases} $$

3. Simplify the Profit Function

$$ P(x) = \begin{cases} x, & \text{if } D \geq x \

1. 75D + 0.50x - 0.50D - 0.75x, & \text{if } D < x \end{cases} $$

$$ P(x) = \begin{cases} x, & \text{if } D \geq x \

1. 25D - 0.25x, & \text{if } D < x \end{cases} $$

2. Expected Profit Calculation

The expected profit $E[P(x)]$ is calculated by integrating over the possible demand values, considering the probability density function of the uniform distribution. The probability density function $f(D)$ for a uniform distribution between 60 and 160 is:

$$ f(D) = \frac{1}{160 - 60} = \frac{1}{100}, \quad 60 \leq D \leq 160 $$

Now, we compute the expected profit:

$$ E[P(x)] = \int_{60}^{x} (1.25D - 0.25x)f(D) , dD + \int_{x}^{160} x f(D) , dD $$

Plug in $f(D) = \frac{1}{100}$:

$$ E[P(x)] = \frac{1}{100} \int_{60}^{x} (1.25D - 0.25x) , dD + \frac{1}{100} \int_{x}^{160} x , dD $$

7. Evaluate the Integrals

$$ E[P(x)] = \frac{1}{100} \left[ 1.25 \frac{D^2}{2} - 0.25xD \right]{60}^{x} + \frac{1}{100} \left[ xD \right]{x}^{160} $$

$$ E[P(x)] = \frac{1}{100} \left[ 0.625D^2 - 0.25xD \right]{60}^{x} + \frac{1}{100} \left[ xD \right]{x}^{160} $$

$$ E[P(x)] = \frac{1}{100} \left[ (0.625x^2 - 0.25x^2) - (0.625(60)^2 - 0.25x(60)) \right] + \frac{1}{100} \left[ 160x - x^2 \right] $$

$$ E[P(x)] = \frac{1}{100} \left[ 0.375x^2 - (0.625(3600) - 15x) \right] + \frac{1}{100} \left[ 160x - x^2 \right] $$

$$ E[P(x)] = \frac{1}{100} \left[ 0.375x^2 - 2250 + 15x + 160x - x^2 \right] $$

$$ E[P(x)] = \frac{1}{100} \left[ -0.625x^2 + 175x - 2250 \right] $$

8. Optimize the Expected Profit

To maximize the expected profit, take the derivative of $E[P(x)]$ with respect to $x$ and set it equal to zero:

$$ \frac{dE[P(x)]}{dx} = \frac{1}{100} \left[ -1.25x + 175 \right] = 0 $$

$$ -1.25x + 175 = 0 $$

$$

1. 25x = 175 $$

$$ x = \frac{175}{1.25} = 140 $$

9. Consider the Constraint In this case, the optimal number of chicken tacos to prepare is 140.