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Suppose that the daily demand (𝐷) for a certain item in the capital market is random and behaves according to a uniform probability (quantity) law with 𝑑 = 0.1, … ,19. Every day, 𝑥 units of the item are brought for sale, which are sold at 5 mu per unit and, if not sold, 2 mu are lost (for storage, return or other). Determine the number of these items, which should be brought at the beginning of the day, so that the expected utility is maximum.

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Answer to a math question Suppose that the daily demand (𝐷) for a certain item in the capital market is random and behaves according to a uniform probability (quantity) law with 𝑑 = 0.1, … ,19. Every day, 𝑥 units of the item are brought for sale, which are sold at 5 mu per unit and, if not sold, 2 mu are lost (for storage, return or other). Determine the number of these items, which should be brought at the beginning of the day, so that the expected utility is maximum.

Expert avatar
Murray
4.5
90 Answers
\text{Expected Profit} = \frac{\int_{0.1}^{19} \left[ \min(5x, 7D - 2x) \right] \, dD}{19 - 0.1}

Since there's a uniform distribution, we consider the two cases where the profit equations exist:

1. **Demand \( D \geq x \) :**

\text{Profit} = 5x

2. **Demand \( D < x \) :**

\text{Profit} = 7D - 2x

Integrate over both segments accordingly:

E[\text{Profit}] = \int_{0.1}^{x} \left( 7D - 2x \right) \frac{dD}{18.9} + \int_{x}^{19} 5x \frac{dD}{18.9}

Solve these integrals:

E[\text{Profit}] = \left( \frac{7}{18.9} \int_{0.1}^{x} D \, dD - \frac{2x}{18.9} \int_{0.1}^{x} dD \right) + \frac{5x}{18.9} \int_{x}^{19} dD

= \frac{7}{18.9} \cdot \left[ \frac{D^2}{2} \right]_{0.1}^{x} - \frac{2x}{18.9} \left[ D \right]_{0.1}^{x} + \frac{5x}{18.9} \left[ D \right]_{x}^{19}

Evaluating these:

1. First part:

= \frac{7}{18.9} \left( \frac{x^2}{2} - \frac{0.1^2}{2} \right) - \frac{2x}{18.9} (x - 0.1)

2. Second part:

= \frac{5x}{18.9} (19 - x)

Combine and simplify:

E[\text{Profit}] = \frac{7x^2 - 7 \cdot 0.01}{2 \times 18.9} - \frac{2x^2 - 0.2x}{18.9} + \frac{5x (19 - x)}{18.9}

After simplifying, the expected profit equation in terms of \( x \):

E[\text{Profit}] = \frac{(7x^2 - 0.07) - (2x^2 - 0.2x) + 5x(19 - x)}{18.9}

Set the derivative of accumulated profit with respect to \( x \) to 0 and solve for \( x \):

\frac{d}{dx} \left[ \frac{5x (19 - x) + 7 \frac{x^2}{2} - 0.07 - 2x^2 + 0.2x}{18.9} \right] = 0

\frac{d}{dx} \left[ \frac{13.6x - \frac{7x^2}{2} - 0.07 - 2x^2 + 0.2x}{18.9} \right] =0

13.6 \quad \text{results in maximum expected profit}

Therefore:

x = 13.6

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