AIM: How do we use derivatives to solve Optimization problems?

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Presentation transcript:

AIM: How do we use derivatives to solve Optimization problems? December 5, 2012 AIM: How do we use derivatives to solve Optimization problems? Do Now: When does the bug change direction? When is the bug’s speed the greatest? HW3.7 Pg. 223 – 225 # 17, 19, 23, 27, 33, 35

What are Optimization Problems? Word Problems! - Real world scenarios that ask you to find the maximum or minimum value! Typical situations include… - Find the route that will take you the shortest time to walk to school. Build a solid using the least amount of material. Build a solid containing the maximum volume. Find the maximum area of a field with fencing. Find the maximum height of a ball thrown in the air.

Example 1 Two numbers sum up to 40. Find the two numbers that maximum their product. Step 1: Read the problem carefully & draw a picture Step 2: Write the equation that needs to be optimized. Step 3: Write a second equation if your original equation includes more than a single variable – a constraint/restriction Step 4: Using the ‘function’ that must be optimized, take derivative and set it equal to zero. Step 5: Use the value to answer the question properly!

Example 2 Tracy wants to make the biggest garden possible, but only has 50 feet of fencing. What are the dimensions that will give her a maximum area?

Remember, you are finding the max and mins! When Optimizing… Remember, you are finding the max and mins! Optimizing on an open interval Find derivative and set it equal to zero. (identify your critical points) Create a number line, verify it is max/min. [First Derivative Test!] OR – do the second derivative test (2) Optimizing on a closed interval Find your critical points. Plug points into f(x) and also test your endpoints! The largest and smallest values are your extrema!

Several Important Equations Area of a rectangle = length x width Perimeter of a rectangle = 2l + 2w Distance = rate x time Volume of a rectangular prism = lwh Volume of cylinder = 𝜋 𝑟 2 ℎ Surface Area of Cylinder = 2𝜋 𝑟 2 + 2𝜋𝑟ℎ