1. 6.2 – Optimization using Trigonometry
IB Math SL/HL, MCB4U - Santowski

2. (A) Review
When a function has a local max/min at x = c, then the value of the derivative is 0 at c
Alternatively, some functions will have local max/min points, even though the derivative is not zero  i.e. the absolute value function (which will have an undefined derivative at the point of the max/min)
However, the converse  when a function has a zero first derivative, the critical point may or may not be an extreme value (i.e. max/min)
So we need to “test” whether the critical point is a max or a min or neither  thus we have either our first derivative test or our second derivative test
In the first derivative test, we look for sign changes in the derivative values before and after the critical point  if the derivative changes from +ve to –ve, then the fcn has a max and vice versa for a min
In the second derivative test, we test for concavity  a fcn will have a max only if the curve is concave down, so if the second derivative is negative, then the critical point is a maximum and vice versa for a min
The maximum and minimum points and values have important roles to play in mathematical modelling.
In optimizing problems, we try to solve problems involving maximizing areas, volumes, profits and minimizing distances, times, and costs

3. (B) Examples
Find the maximum perimeter of a right triangle with a hypotenuse of 20 cm
We can set up a Pythagorean relationship (P = x (400 – x2) )
Or we can draw a right triangle, set  as the base angle  then the adjacent side measures 20cos and the opposite side measures 20sin and of course the hypotenuse is 20 cm
p() = 20cos + 20sin + 20
Then p`() = -20sin() + 20cos() = 0
So sin() = cos() or 1 = tan()    = /4 or 45°
Then p(/4) = 20(1 + cos(/4) + sin(/4)) = 2 = 48.3 cm
To verify a maximum, we can take the second derivative 
p``() = -20cos() – 20 sin()
p``(/4) = -20cos(/4) – 20 sin(/4) = -40/2
So our second derivative is –ve meaning the function curve is concave down, meaning we have a maximum point at /4 giving 48.3 as the max perimeter

4. (C) Examples
Ex 2. Find the area of the largest rectangle that can be inscribed in a semicircle of radius r
Show how to develop the formula:
A() = 2rcos()  rsin()

5. (D) Internet Links
Optimization from Calculus 1 - Problems and Solutions from Pheng Kim Ving
Maximum/Minimum Problems and Solutions from UC Davis
Calculus I (Math 2413) - Applications of Derivatives - Optimization from Paul Dawkins
Optimization Problems and Applets from Visual Calculus

6. (E) Homework Stewart, 1989, Chap 7.4, p325, Q3-8
Photocopy from Stewart, 1998, Chap 4.6, p318, Q22-33