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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 prob

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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.

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84 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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