Challenging Real-World Math Problems for 16-Year-Old Students

Instructions

Solve the following problems, showing all of your work. These problems are designed to be challenging and may require combining concepts from algebra, geometry, and pre-calculus. Partial credit may be given for logical reasoning and correct steps, even if the final answer is incorrect. Calculators are permitted, but all formulas and setup must be shown.

Problem 1: The Retirement Portfolio Puzzle

An investor begins saving for retirement at age 25 and plans to retire at age 65 (a 40-year investment period). They contribute $6,000 at the end of each year. They are considering two investment strategies:

• Option A: A conservative fund with a 5% annual interest rate, compounded continuously.
• Option B: A stock market fund with an 8% annual interest rate, compounded monthly. However, this fund is riskier. At age 50 (after 25 years of investing), a one-time market crash occurs, instantly reducing the account's total value by 30%. After the crash, the account continues to grow at 8% compounded monthly, and the investor continues their $6,000 annual contributions.

Which option results in a higher account balance at age 65? By how much more, to the nearest dollar?

Problem 2: The Optimal Drone Delivery Route

A delivery company uses a drone to drop off packages. The drone starts at the central warehouse, W(0, 0), must visit three locations, A(8, 12), B(16, 4), and C(2, 7), and then return to the warehouse. All coordinates are in kilometers.

• The drone's battery has a maximum range of 55 km for a single trip.
• There is a circular no-fly zone centered at P(9, 7) with a radius of 2.5 km. The drone flies in straight lines between its stops, and its flight path cannot enter this circle.

Your task is to find the shortest possible route for the drone that visits all three locations. Then, determine if this optimal route is feasible. To be feasible, the route must satisfy two conditions:

1. The total distance must not exceed the drone's 55 km range.
2. The flight path must not intersect the no-fly zone.

Problem 3: The Pharmaceutical Production Plan

A pharmaceutical company manufactures two medications, Drug X and Drug Y. The production of each drug requires three chemical precursors: A, B, and C. The company has a limited supply of these precursors.

• Available Supply: 1200g of Precursor A, 1800g of Precursor B, and 1500g of Precursor C.
• Requirements per Batch:
  • One batch of Drug X requires: 10g of A, 30g of B, and 10g of C.
  • One batch of Drug Y requires: 20g of A, 20g of B, and 30g of C.
• Profit: The company makes a profit of $400 per batch of Drug X and $500 per batch of Drug Y.

How many batches of Drug X and Drug Y should the company produce to maximize its profit, assuming they can produce fractional batches? What is the maximum possible profit? (For an extra challenge, determine the optimal number of whole batches and the corresponding profit).

Answer Key

Problem 1: The Retirement Portfolio Puzzle

This problem involves calculating the future value of a series of investments (an annuity).

Option A (Continuous Compounding):

The future value of each annual $6,000 contribution (C) made at the end of year k is C * e^r(T-k), where r=0.05 and T=40 years. We must sum this for k=1 to 40. This forms a geometric series. The sum (S) can be calculated as:
S = C * (e^rT - 1) / (e^r - 1)
S = 6000 * (e^{(0.05 * 40)} - 1) / (e^0.05 - 1)
S = 6000 * (e² - 1) / (e^0.05 - 1)
S = 6000 * (7.389056 - 1) / (1.051271 - 1)
S = 6000 * (6.389056) / (0.051271) ≈ $747,714

Option B (Monthly Compounding with a Crash):

This must be calculated in two stages.

Stage 1: First 25 years (Age 25 to 50)
We calculate the future value of a series of 25 annual payments in an account compounding monthly. We use the effective annual rate (EAR) to simplify: EAR = (1 + r/n)ⁿ - 1 = (1 + 0.08/12)¹² - 1 ≈ 8.30%. The future value factor for the annuity is ((1+EAR)^t - 1) / EAR.
Value at age 50 = 6000 * [((1.083)²⁵ - 1) / 0.083]
Value at age 50 ≈ 6000 * [(7.984 - 1) / 0.083] ≈ $492,819

Market Crash: The balance is reduced by 30%.
Balance after crash = $492,819 * (1 - 0.30) = $344,973.30

Stage 2: Next 15 years (Age 50 to 65)
First, the crashed balance grows as a lump sum for 15 years:
FV_{lump sum} = $344,973.30 * (1 + 0.08/12)^(12*15) ≈ $344,973.30 * 3.306 ≈ $1,140,491
Second, calculate the future value of the next 15 years of contributions:
FV_annuity = 6000 * [((1.083)¹⁵ - 1) / 0.083] ≈ 6000 * [(3.39 - 1) / 0.083] ≈ $172,771

Total for Option B:
Total = FV_{lump sum} + FV_annuity = $1,140,491 + $172,771 = $1,313,262

Conclusion:
Option B yields a higher balance.
Difference = $1,313,262 - $747,714 = $565,548

Problem 2: The Optimal Drone Delivery Route

Step 1: Calculate Distances and Find the Shortest Route

Using the distance formula d = √((x₂-x₁)² + (y₂-y₁)²):

• d(W,A) = √(8²+12²) = √208 ≈ 14.42 km
• d(W,B) = √(16²+4²) = √272 ≈ 16.49 km
• d(W,C) = √(2²+7²) = √53 ≈ 7.28 km
• d(A,B) = √((16-8)²+(4-12)²) = √128 ≈ 11.31 km
• d(A,C) = √((8-2)²+(12-7)²) = √61 ≈ 7.81 km
• d(B,C) = √((16-2)²+(4-7)²) = √205 ≈ 14.32 km

Possible total route distances:

• W-A-B-C-W = 14.42 + 11.31 + 14.32 + 7.28 = 47.33 km
• W-A-C-B-W = 14.42 + 7.81 + 14.32 + 16.49 = 53.04 km
• W-C-A-B-W = 7.28 + 7.81 + 11.31 + 16.49 = 42.89 km (Shortest)
• The reverse path, W-B-A-C-W, has the same length.

The shortest route is approximately 42.89 km. This is within the 55 km battery limit.

Step 2: Check for No-Fly Zone Intersection

The no-fly zone is a circle: (x - 9)² + (y - 7)² = 2.5² = 6.25.
We must check if any segment of the shortest path (W-C-A-B-W) gets closer than 2.5 km to the center P(9,7).

• Path W(0,0) to C(2,7): The center of the circle is P(9,7). The y-coordinate is the same as C. The closest point on the line WC to P is not on the segment WC. The closest points on the segment are the endpoints.
  d(P,C) = √((9-2)² + (7-7)²) = 7 km. Safe.
• Path C(2,7) to A(8,12): Line equation: 5x - 6y + 32 = 0.
  Distance from P(9,7) to line = |5(9) - 6(7) + 32| / √(5² + (-6)²) = |35| / √61 ≈ 4.48 km. Safe.
• Path A(8,12) to B(16,4): Line equation: x + y - 20 = 0.
  Distance from P(9,7) to line = |9 + 7 - 20| / √(1² + 1²) = |-4| / √2 ≈ 2.82 km. This is very close, but 2.82 > 2.5. Safe.
• Path B(16,4) to W(0,0): Line equation: x - 4y = 0.
  Distance from P(9,7) to line = |9 - 4(7)| / √(1² + (-4)²) = |-19| / √17 ≈ 4.61 km. Safe.

Conclusion:
Yes, the optimal route is feasible. The shortest path is W-C-A-B-W (or reverse), with a length of 42.89 km, which is within the 55 km battery limit. All segments of this path remain more than 2.5 km away from the center of the no-fly zone.

Problem 3: The Pharmaceutical Production Plan

Step 1: Define Variables and Constraints

Let x = batches of Drug X, y = batches of Drug Y.

• A: 10x + 20y ≤ 1200 => x + 2y ≤ 120
• B: 30x + 20y ≤ 1800 => 3x + 2y ≤ 180
• C: 10x + 30y ≤ 1500 => x + 3y ≤ 150
• x ≥ 0, y ≥ 0

Step 2: Define Objective Function

Profit P = 400x + 500y. We must maximize this.

Step 3: Find Vertices of the Feasible Region

By graphing the inequalities, we find the region of possible (x, y) values. The maximum profit will occur at one of the vertices (corners) of this region.

• Vertex 1: (0, 0) - Origin
• Vertex 2: (0, 50) - y-intercept of x + 3y = 150
• Vertex 3: (60, 0) - x-intercept of 3x + 2y = 180
• Vertex 4: Intersection of 3x + 2y = 180 and x + 3y = 150.
  Solving the system gives x = 240/7 and y = 270/7.
  Vertex is (²⁴⁰⁄₇, ²⁷⁰⁄₇) ≈ (34.28, 38.57).

(Note: The constraint x + 2y ≤ 120 is redundant as the other constraints are stricter in the feasible region.)

Step 4: Test Vertices in the Profit Function

• P(0, 0) = $0
• P(0, 50) = 400(0) + 500(50) = $25,000
• P(60, 0) = 400(60) + 500(0) = $24,000
• P(²⁴⁰⁄₇, ²⁷⁰⁄₇) = 400(²⁴⁰⁄₇) + 500(²⁷⁰⁄₇) = (96000 + 135000) / 7 = 231000 / 7 ≈ $33,000

Conclusion (Fractional Batches):

To maximize profit, the company should produce ²⁴⁰⁄₇ (approx. 34.28) batches of Drug X and ²⁷⁰⁄₇ (approx. 38.57) batches of Drug Y. The maximum possible profit is $33,000.

Challenge (Whole Batches):

Since batches must be whole numbers, we test integer points near the optimal vertex (34.28, 38.57) that are within the feasible region. The best integer point is (33, 39).

• Check (33, 39):
  • A: 10(33) + 20(39) = 1110 ≤ 1200 (Ok)
  • B: 30(33) + 20(39) = 1770 ≤ 1800 (Ok)
  • C: 10(33) + 30(39) = 1500 ≤ 1500 (Ok)
• Profit P(33, 39) = 400(33) + 500(39) = 13,200 + 19,500 = $32,700.

The optimal integer solution is to produce 33 batches of Drug X and 39 batches of Drug Y for a maximum profit of $32,700.