Problem Solving > Optimization Problems

**Optimization problems** in calculus often involve the determination of the “optimal” (meaning, the best) value of a quantity. For instance, we might want to know the biggest area that a piece of rope could be tied around. Or how high a ball could go before it falls back to the ground. Or at which point of a loop does a roller coaster run the slowest. Very often, the optimization must be done with certain **constraints**. In the case of the rope, we’re limited by its length. These constraints are usually very helpful to solve optimization problems. Optimal values are often either the **maximum **or the** minimum** values of a certain function.

## Optimization Problems in Calculus: Steps.

Sample problem: Find the maximum area of a rectangle whose perimeter is 100 meters. (Note: This is a typical optimization problem in **AP calculus**).

Step 1:** Determine the function** that you need to optimize. In the sample problem, we need to optimize the area A of a rectangle, which is the product of its length L and width W. Our function in this example is A = LW.

Step 2:** Identify the constraints** to the optimization problem. In our sample problem, the perimeter of the rectangle must be 100 meters. This will be useful in the next step.

Step 3: Express that function in terms of a **single variable **upon which it depends, using algebra. For this example, we’re going to express the function in a single variable. “L.”

- A rectangle’s perimeter is the sum of its sides, that is, 100m = 2L + 2W.
- Subtract 2L from both sides of this equation, 2W = 100m – 2L.
- Divide each side by 2: W = 50m – L.
- Substitute 50m – L for “W” in A = LW: A = L (50m – L) = 50m L – L².

Step 4: Calculate the derivative of the function with respect to a variable. The derivative dA/dL = 50m (1) L^{(1-1)} – 2 L^{(2-1)} = 50m – 2L.

Step 5:** Set the function to zero** and compute the corresponding variable’s value. For our sample problem, we set dA/dL = 0 = 50m – 2L. So L = 25m.

Step 6: Use the value from Step 5 to calculate the corresponding **optimal value** of the function. In our sample problem, A = 50m L – L² = 50 m (25m) – (25m)² = 625 m².

*That’s how to solve optimization problems in calculus!*