Solving Optimization Problems with TI-84 Calculator

Introduction

Optimization problems are among the most practical and frequently tested applications of calculus and algebra. They require you to find the greatest or least possible value of a quantity under specific constraints — such as maximizing profit, maximizing enclosed volume or area, minimizing material costs, or minimizing travel time.

While optimization can be solved algebraically using derivatives, the TI-84 calculator dramatically accelerates the process by graphing objective functions and locating exact maximum and minimum coordinates. Practice these problems directly on our free browser-based TI-84 Calculator Online.

The 6-Step Optimization Framework

1. Identify Variables: Clearly define what you are optimizing (e.g. (A) for Area, (P) for Profit) and your independent variable (e.g. (x)).
2. Write the Constraint Equation: Formulate the equation representing physical limitations (e.g. total fencing perimeter or material budget).
3. Formulate the Objective Function: Express the quantity to be optimized in terms of a single variable.
4. Graph on the TI-84: Enter the objective function into [Y=] and press [GRAPH].
5. Find Extrema: Use [2ND] + [TRACE] > 4:maximum or 3:minimum to calculate the critical vertex.
6. Interpret Results: State your answer with appropriate units in the context of the question.

Worked Example 1: Maximum Enclosed Area

Problem: A farmer has 100 meters of fencing to build a rectangular pasture. What dimensions produce the maximum area?

• Constraint: Perimeter (2L + 2W = 100 implies W = 50 – L).
• Objective Function: (text{Area} = L times W = L(50 – L) = 50L – L^2).
• Calculator Steps:
  1. Press [Y=] and enter 50X - X² into Y1.
  2. Press [WINDOW]: Set Xmin = 0, Xmax = 50, Ymin = 0, Ymax = 700.
  3. Press [GRAPH].
  4. Press [2ND] > [TRACE] > 4:maximum.
  5. Select Left Bound at (x = 20), Right Bound at (x = 30), and press [ENTER] on Guess.
• Result: (X = 25, Y = 625). The maximum area is 625 m² with dimensions 25 m × 25 m (a perfect square).

Worked Example 2: Maximum Business Profit

Problem: A company’s profit in dollars is modeled by (P(x) = -2x^2 + 120x – 500), where (x) is units sold. Find maximum profit.

• Enter -2X² + 120X - 500 into [Y=].
• Press [2ND] > [TRACE] > 4:maximum.
• Set bounds around the curve’s peak.
• Result: (X = 30, Y = 1300). Maximum profit is $1,300 achieved when selling 30 units.

Common Mistakes to Avoid

• Graphing the Constraint Instead of the Objective: Always graph the function you want to maximize or minimize (Profit, Area, Cost), not the perimeter or constraint equation.
• Window Bounds Too Narrow: If the peak isn’t visible, press [ZOOM] > 0:ZoomFit or manually increase Ymax in [WINDOW].
• Forgetting to State Both Dimensions: Remember that (x) gives the dimension, and (y) gives the optimal area/profit. Read the question carefully to provide what is asked.

Frequently Asked Questions

Can I verify optimization results using table values?

Yes. Press [2ND] > [GRAPH] to view the numerical table. The peak value will have smaller outputs on both sides. Also review our guide on Finding Maximum & Minimum Values.

Summary

The TI-84 turns complex calculus optimization into an intuitive, visual workflow. Practice formulating and solving optimization models today on the live TI-84 simulator!