The Optimization Station
Below you will find the end-all, be-all format for solving optimization problems in the great course of AP Calculus. Adhere strictly to these rules for optimization problems for the rest of your life.
Before we begin, keep in mind that when solving optimization problems, you are looking for ABSOLUTE MAXIMUM OR ABSOLUTE MINIMUM values.
- Determine what you are trying to maximize or minimize in the problem.
□Write down all important information for the problem
□Draw a diagram of the situation if possible
- Write the PRIMARY EQUATION for the problem (the primary equation is the equation which represents the quantity we are trying to maximize or minimize)
□If the primary equation has more than one independent variable, use substitution to replace any ‘extra’ independent variables in the equation. Remember, you want to have ONLY ONE independent variable.
- Determine the DOMAIN for the primary equation.
□Pay attention to the problem situation…where does it make sense for the independent variable to exist?
□Visualize what will happen if the independent variable gets large or small. Can it be negative? What is the largest possible value for the variable?
- OPTIMIZATION: Now, you will begin the key step in the entire process. Perform the following necessary steps to find the absolute extrema, should they exist:
□Differentiate the primary equation
□Find the critical values of the primary equation by determining where its derivative equals zero or where it does not exist
□PIVOT STEP #1: Should you choose this step, you will find maximum or minimum values using the FIRST DERIVATIVE TEST (that is, test the derivative to determine where it is positive and where it is negative). Relative maximum values will occur wherever the derivative switches from positive to negative; relative minimum values will occur wherever the derivative switches from negative to positive.
□PIVOT STEP #2: Should you choose this step, you will find maximum or minimum values using the SECOND DERIVATIVE TEST (that is, test values where the first derivative equals zero with the second derivative). If the second derivative is positive, a relative minimum exists; if the second derivative is negative, then a relative maximum exists.
- JUSTIFICATION: In this step, you must carefully choose how to JUSTIFY your answer. Your justification depends upon whether or not the domain of the primary equation is an open, infinite interval or a closed interval. Refer to the chart below for deciding how to justify:
INTERVAL HAS ENDPOINTS / INTERVAL IS OPEN
□After finding the relative max/min values, justify why it is a local min or max (derivative changes sign, or 2nd derivative is positive or negative, etc.)
□Test the critical values and the endpoints with the primary equation
□Show each test by either listing them in succession or by using a chart
□State your conclusion from the test of the critical values and the endpoints / □After finding the relative max/min values, justify why it is a local min or max (derivative changes sign, or 2nd derivative is positive or negative, etc.)
□If you have only one critical value on the interval, is must yield either the absolute max or absolute min
□You will reason this by either analyzing the first derivative (state behavior or function based on its increasing/decreasing nature) or analyzing the second derivative (verifying the concavity of the function)
- ANSWER THE QUESTION. ANSWER THE QUESTION. ANSWER THE QUESTION. ANSWER THE QUESTION. ANSWER THE QUESTION. ANSWER THE QUESTION.
A USEFUL THEOREM
Let f be continuous on an interval I and assume that f has exactly one relative extremum on I, say at c.
□If f has a relative minimum at c, then f(c) is the absolute minimum value on f on the interval I.
□If f has a relative maximum at c, then f(c) is the absolute maximum value on f on the interval I.
COMMENTS:
□Write your justifications using complete sentences. If you are lazy, your points will become ‘hazy.’
□Answer the question using a complete sentence.
□Be sure to include the correct units with all answers.
Sample #1 – ENDPOINTS
An open box is to be made from a 3’ x 8’ piece of sheet metal by cutting out squares of equal size from the four corners and bending up the sides. Find the maximum volume that the box can have.
Sample #2 – NO ENDPOINTS
A closed cylindrical can is to hold 1000 cm3 of liquid. What dimensions for the height and radius will minimize the amount of material needed to manufacture the can?
Optimization Station – Do not lose the sheet. EVER.1