Calculus 1 -- Optimization -- Practice

0:00 Introduction
0:56 Problem 1
9:06 Problem 2
18:40 Problem 3
26:00 Problem 4
39:28 Problem 5
51:30 Problem 6
1:01:08 Problem 7

[jana]

I will provide 2 alternate solutions for problem 1:

Solution 1: note that x + 2y = 400, and x, y are positive since it is a fence. Thus, by AM-GM we have 200 ≥ √(2xy). Squaring both sides gives us 2 * 100^2 ≥ xy, which implies the maximum area is 2 * 100^2.

Solution 2: note that x + 2y = 400, and x = 400 - 2y. Area is xy = 400y - 2y^2, completing the square we have -2(y-100)^2 + 100^2 * 2. Thus, the maximum is achieved at y = 100 and that implies x = 200. Therefore, the max area is 200 * 100.

[osasikemwenogieva]

Thank you, Professor Butler. You are a life saver.