Examples: 1. Minimization A wholesaler is to transport 840 crates of soda from the factory to his shop. He has a lorry which can carry 150 crates at a time and a pick-up which can carry 60 crates at a time. The cost of each journey for the lorry is Shs 25,000 and for the pick-up, it is Shs 20,000. Due to time constraint, the pick-up is to make more journeys than the lorry because it travels faster. He has Shs 250,000 available for transportation. i. As a consultant, advise how he can organize his vehicles to keep cost as low as possible and state how much he saves by making these journeys. ii. Suppose there was no time constraint, determine how he could have organized his vehicles to minimize the transportation cost and state how much he would save.

To solve this minimization problem, we will break down the two scenarios provided and use a combination of algebra and optimization techniques to find the solution.

### Definitions and Constraints
Let's define:
- \( x \): the number of trips made by the lorry
- \( y \): the number of trips made by the pick-up

From the problem, we have the following information:
- The lorry can carry 150 crates per trip.
- The pick-up can carry 60 crates per trip.
- The total crates to be transported is 840.
- The cost per trip for the lorry is Shs 25,000.
- The cost per trip for the pick-up is Shs 20,000.
- Available budget: Shs 250,000.
- The pick-up is to make more journeys than the lorry (time constraint in scenario (i)).

#### Constraints:
1. **Transportation Capacity**: The lorry and the pick-up must together transport a total of 840 crates.
   \[
   150x + 60y = 840
   \]
   
2. **Cost Constraint**: The cost for the journeys should not exceed the budget.
   \[
   25000x + 20000y \leq 250000
   \]
   
3. **Time Constraint (Scenario i)**: The pick-up makes more trips than the lorry.
   \[
   y > x
   \]

### Objective
We need to minimize the cost function:
\[
\text{Cost} = 25000x + 20000y
\]

### Solution

#### (i) With Time Constraint (Pick-up makes more trips than the lorry)

1. **Solving the Transportation Constraint**: 
   From \( 150x + 60y = 840 \), simplify by dividing by 30:
   \[
   5x + 2y = 28
   \]
   
   Rearrange for \( y \):
   \[
   y = \frac{28 - 5x}{2}
   \]
   Since \( y \) and \( x \) represent trips, they must be non-negative integers. Additionally, \( y > x \) (time constraint).

2. **Substitute into Cost Function**:
   The goal is to minimize \( 25000x + 20000y \) under the budget constraint \( 25000x + 20000y \leq 250000 \).

3. **Finding Integer Values**:
   By testing integer values of \( x \) that satisfy both \( y > x \) and the budget constraint, we find the smallest cost configuration under the time constraint.

#### (ii) Without Time Constraint

In this scenario, the only constraints are the transportation capacity and the budget. We can minimize the cost function by allowing any values of \( x \) and \( y \) that satisfy:
   \[
   5x + 2y = 28
   \]
and the budget:
   \[
   25000x + 20000y \leq 250000
   \]
We find the smallest integer values of \( x \) and \( y \) that minimize cost.

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### Further Questions:
1. What if the cost of each trip for the lorry and pick-up changed, how would this affect the solution?
2. How would you approach this if the budget constraint was removed entirely?
3. What if the transportation requirement was increased to 1,000 crates, how would the strategy change?
4. Could this problem be solved using linear programming methods? If so, how?
5. How could the addition of a second lorry impact the minimization strategy?

### Tip:
When facing minimization problems, always start by defining the constraints clearly before moving to optimization—this ensures feasible solutions align with all requirements.

A wholesaler is to transport 840 crates of soda from the factory to his shop. He has a lorry which can carry 150 crates at a time and a pick-up which can carry 60 crates at a time. The cost of each journey for the lorry is Shs 25,000 and for the pick-up, it is Shs 20,000. Due to time constraint, the pick-up is to make more journeys than the lorry because it travels faster. He has Shs 250,000 available for transportation. i. As a consultant, advise how he can organize his vehicles to keep cost as low as possible and state how much he saves by making these journeys. ii. Suppose there was no time constraint, determine how he could have organized his vehicles to minimize the transportation cost and state how much he would save.

Learn how to minimize transportation costs for a wholesaler who needs to transport 840 crates of soda. The solution involves optimization techniques, including handling time constraints and budget limitations for the lorry and pick-up truck. Ideal for high school students in Grades 11-12 studying optimization problems in mathematics.

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