Math Problem Statement

We have a monthly budget of $7160 and three types of developers with different daily rates. We need to determine the maximum number of days each developer can work within this budget, considering the constraint that TAs cannot work more than 3 days. We cannot have fractional days. We can no allocate any developer more than 20 days Sr Dev : $224/day TA : $280/day Dev: $192/day

Solution

To determine the maximum number of days each developer can work within the budget, we need to solve a linear optimization problem. We have three types of developers, each with a different daily rate, and we are given the constraints:

  • Sr Dev: $224/day
  • TA: $280/day (with a maximum of 3 days)
  • Dev: $192/day

We need to find the maximum number of days each developer can work without exceeding the total monthly budget of $7160. Additionally, no developer can work more than 20 days.

Let's define:

  • x1x_1 = number of days the Sr Dev works
  • x2x_2 = number of days the TA works (with a maximum of 3 days)
  • x3x_3 = number of days the Dev works

Objective:

Maximize the total number of working days: x1+x2+x3x_1 + x_2 + x_3

Constraints:

  1. Budget constraint: 224x1+280x2+192x37160224x_1 + 280x_2 + 192x_3 \leq 7160
  2. Maximum working days for each developer: x120x_1 \leq 20, x23x_2 \leq 3, x320x_3 \leq 20
  3. Non-negativity: x10x_1 \geq 0, x20x_2 \geq 0, x30x_3 \geq 0

We can solve this by testing the maximum allowable days within these constraints.

Let's proceed step by step:

  1. Set x2x_2 to 3 (the maximum allowable days for TA) because TA has the highest rate and is limited.
  2. Maximize x1x_1 and x3x_3 within the remaining budget.

Let's calculate:

  • For x2=3x_2 = 3:

    • TA cost: 280×3=840280 \times 3 = 840
    • Remaining budget: 7160840=63207160 - 840 = 6320
  • Maximize x1+x3x_1 + x_3 under the new budget constraint:

    • 224x1+192x36320224x_1 + 192x_3 \leq 6320

Now, we check different combinations for x1x_1 and x3x_3:

If x1=20x_1 = 20, then:

  • Cost of Sr Dev: 224×20=4480224 \times 20 = 4480
  • Remaining budget: 63204480=18406320 - 4480 = 1840
  • Maximum x3x_3 given remaining budget: 18401929.58\frac{1840}{192} \approx 9.58 (rounded down to 9 days)

Let's verify:

  • Total days:
    • x1=20x_1 = 20, x2=3x_2 = 3, x3=9x_3 = 9
  • Total cost:
    • 224×20+280×3+192×9=4480+840+1728=7048224 \times 20 + 280 \times 3 + 192 \times 9 = 4480 + 840 + 1728 = 7048

This fits within the budget, and it's the maximum combination.

Solution:

  • Sr Dev: 20 days
  • TA: 3 days
  • Dev: 9 days

This allocation will utilize $7048 of the $7160 budget.

Would you like any further details or have any questions?

Relative Questions:

  1. How would the solution change if the budget increased?
  2. What if TAs could work more than 3 days?
  3. How would changing the daily rate of one of the developers impact the solution?
  4. What happens if we require that each developer works at least 5 days?
  5. How can we optimize the days if fractional days were allowed?

Tip:

When dealing with linear optimization problems, identifying constraints and systematically testing possible combinations can lead to an optimal solution.