a gas station sells Q gallons of gasoline per year, which is delivered N times per year in equal shipments of Q/N gallons. The cost if each delivery is d dollars and yearly storage costs are sQT where T is the length of time(fraction of year) between shipments and s is a constant. Which value of N minimise the cost. Express T in terms of N

To solve this problem, we need to first express the total cost \( C \) as a function of \( N \), the number of deliveries per year. The total cost consists of the delivery cost and the storage cost.

### Step 1: Expressing the Delivery Cost

The delivery cost per year is simply the number of deliveries \( N \) multiplied by the cost per delivery \( d \):

\[ \text{Delivery Cost} = N \cdot d \]

### Step 2: Expressing the Storage Cost

To find the storage cost, we first need to determine the average amount of gasoline stored. Since each delivery is \( \frac{Q}{N} \) gallons, and the gasoline is used evenly throughout the year, the average amount stored is \( \frac{Q}{2N} \) gallons.

Next, we need to express the length of time \( T \) between shipments in terms of \( N \). Since there are \( N \) deliveries per year, the time between each delivery is:

\[ T = \frac{1}{N} \]

The yearly storage cost is given by \( sQT \). Substituting \( T = \frac{1}{N} \), we get:

\[ \text{Yearly Storage Cost} = sQ \left(\frac{1}{N}\right) \]

### Step 3: Total Cost Function

The total cost \( C \) is the sum of the delivery cost and the storage cost:

\[ C = \text{Delivery Cost} + \text{Storage Cost} \]
\[ C = N \cdot d + sQ \left(\frac{1}{N}\right) \]

### Step 4: Finding the Value of \( N \) That Minimizes the Cost

To minimize the total cost \( C \), we take the derivative of \( C \) with respect to \( N \) and set it to zero:

\[ C(N) = Nd + \frac{sQ}{N} \]

\[ \frac{dC}{dN} = d - \frac{sQ}{N^2} \]

Set the derivative equal to zero to find the critical points:

\[ d - \frac{sQ}{N^2} = 0 \]

Solving for \( N \):

\[ d = \frac{sQ}{N^2} \]
\[ N^2 = \frac{sQ}{d} \]
\[ N = \sqrt{\frac{sQ}{d}} \]

### Step 5: Verification

To ensure this critical point is a minimum, we check the second derivative of \( C \):

\[ \frac{d^2C}{dN^2} = \frac{2sQ}{N^3} \]

Since \( \frac{2sQ}{N^3} \) is always positive for positive values of \( s \), \( Q \), and \( N \), the second derivative is positive, indicating a minimum.

### Conclusion

The value of \( N \) that minimizes the total cost is:

\[ N = \sqrt{\frac{sQ}{d}} \]

The length of time \( T \) between shipments, expressed in terms of \( N \), is:

\[ T = \frac{1}{N} = \sqrt{\frac{d}{sQ}} \]

Would you like more details or have any questions?

Here are 5 questions for you to explore:

1. How would changes in the cost per delivery \( d \) affect the optimal number of deliveries \( N \)?
2. What happens to the total cost \( C \) if the yearly storage cost constant \( s \) increases?
3. How can this model be adapted if the gasoline usage is not uniform throughout the year?
4. What are the implications of having very high or very low \( N \)?
5. How can we incorporate varying delivery costs into this model?

**Tip:** Always verify the conditions under which your mathematical models apply, especially when dealing with real-world scenarios.

Learn how to minimize operational costs for a gas station by optimizing delivery frequency. This problem involves deriving and minimizing a total cost function considering delivery costs and storage expenses. Suitable for advanced mathematics enthusiasts and professionals.

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