Question

can you explain optimization problems and the best way to solve them. also if you can explain related rates and the best atrategy. thank you!

Answer 1

Optimization problems will always ask you to maximize or minimize some quantity, having described the situation using words (instead of immediately giving you a function to max/minimize).

How to solve an optimization problem?
1. Step 1: Understand the problem and underline what is important ( what is known, what is unknown,
what we are looking for, dots)
2. Step 2: Draw a “diagram”; if it is possible.
3. Step 3: Assign “symbols” or “variables” for all the quantities involved (know or unknown), and label
the diagram.
4. Step 4: Write the quantity Q to be maximized or minimized in terms of some of the previous variables
(from Step 3). Example: Q = g(x, y, h)
5. Step 5: Rewrite Q as a function of only one variable. To do this, find the relationships between the
variables using the information given in the problem. Then, use these equations to eliminate all but one
of the variables in the expression of Q. Thus, we get Q = f(x).
6. Step 5: Use the methods of sections 10.1 and 10.2 to find the maximum or the minimum of the quantity
Q = f(x).
7. REMARK: Do not forget to find the endpoints and to check if the maximum or the minimum is at these
points if you have more than one critical number in the domain

EXAMPLE- The regular air fare between Boston and San Fransisco is $500. An airline using planes with a capacity of 300passengers on this route observes that they fly with an average of 180 passengers. Market research tells the airlines’ managers that each $ 5 fare reduction would, attract on average, 3 more passengers for each flight. How should they set the fare to maximize their revenue?
• Solution:
Let R= the revenue function = quantity × price
Let n= the number of times the fare is reduced by $ 5 dollars. Then:
( price= $500 − n · ($5) = 500 − 5n dollars,
quantity = number of passengers = 180passengers + n · (3passengers) = 180 + 3n passengers.
Hence R(n) = (180 + 3n)(500 − 5n) = 90 000 + 600n − 15n^2
to maximize (for 0 ≤ n ≤ 40).
R'(n) = 600 − 30n = 0 ⇒ n= 20 is the only critical number.
R"(n) = −30 ⇒ R"(20) = −30 < 0. By the second derivative test, R has a local maximum at n = 20,
which is an absolute maximum since it is the only critical number.
The best fare to maximize the revenue is then: $ 500 − 5(20) = $400 , with 180 + 3(20) = 240 passengers
and a revenue of R(20) = $96, 000 .