Applied Optimization quiz #1 Flashcards
Applied Optimization quiz #1
You can tap to flip the card.
Control buttons has been changed to "navigation" mode.
1/10Skip to main content
What are the general steps to solve an applied optimization problem in calculus?
The steps are: (1) Draw a diagram and define variables, (2) Write a function for the value to optimize in terms of one variable, (3) Determine domain restrictions based on real-world constraints, (4) Find critical points by setting the derivative to zero, and (5) Use the extreme value theorem for closed intervals or the second derivative test for open intervals to identify maximum or minimum values. A coffee shop sells x coffees per day at a price given by p(x) = 100 - x/4. How do you set up the function to maximize the shop's daily revenue, and what are the domain restrictions?
Revenue is R(x) = x * p(x) = x(100 - x/4) = 100x - x^2/4. The domain restrictions are x ≥ 0 (cannot sell negative coffees) and 100 - x/4 ≥ 0 (price cannot be negative), so x ≤ 400. Thus, 0 ≤ x ≤ 400. When optimizing the area of a rectangular fence using 200 feet of fencing with one side against a rock wall, how do you express the area as a function of one variable, and what are the domain restrictions?
Let x be the length of each of the two fenced sides perpendicular to the wall, and y the side parallel to the wall. The constraint is 2x + y = 200, so y = 200 - 2x. The area is A(x) = x * y = x(200 - 2x) = 200x - 2x^2. The domain is 0 ≤ x ≤ 100, since x cannot be negative and 2x cannot exceed 200. What is the first step when solving an applied optimization problem in calculus?
The first step is to draw a diagram and define the variables involved in the problem. How do you express the area of a rectangular fence with one side against a rock wall and 200 feet of fencing as a function of one variable?
Let x be the length of the two sides perpendicular to the wall; then area A(x) = x(200 - 2x) = 200x - 2x^2. What are the domain restrictions for the variable x in the rectangular fence problem with 200 feet of fencing?
The domain is 0 ≤ x ≤ 100, since x cannot be negative and 2x cannot exceed 200. How do you set up the revenue function for a coffee shop selling x coffees per day at a price p(x) = 100 - x/4?
Revenue is R(x) = x * p(x) = 100x - x^2/4. What domain restrictions must be considered when maximizing the coffee shop's daily revenue?
x must satisfy 0 ≤ x ≤ 400, since the number of coffees sold cannot be negative and the price cannot be negative. After finding the critical points in an applied optimization problem, how do you determine if you have a maximum or minimum on a closed interval?
Use the extreme value theorem by evaluating the function at the critical points and endpoints; the largest value is the maximum and the smallest is the minimum. When should you use the second derivative test instead of the extreme value theorem in applied optimization problems?
Use the second derivative test when the domain is an open interval without endpoints; a negative second derivative at a critical point indicates a maximum, and a positive indicates a minimum.
10:13
