Calculus > Optimization Problems in Calculus

**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!*

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That was really simple. I was trying to learn Low thrust trajectory optimizations and was daunted at the very beginning. Thank you for explaining the basics so well

Very well explained in simple terms. Thank you so much.