Blog #19

For this final holiday blog, I’m going to review the last thing we did before exams: Optimization. This is also the hardest thing we’ve done all year (in my opinion). So I’m going to try to simplify it as much as possible, but no guarantees…XD

Okay, optimization is when you are maximizing or minimizing something. The first thing you need to do is read the problem and figure out if you’re minimizing or maximizing. Then, you need to figure out your two equations: primary and secondary equation. Now, your primary equation is the equation used for max/minimizing. And your secondary equation is the equation given to you in the problem. After this, you need to solve your secondary equation for one of the variables. Then, plug your solved secondary equation into your primary equation and simplify. Finally, we begin max/minimizing. First of all, you take the derivative of the equation you just simplified. Then set it equal to zero and solve for x. Next, set up your intervals and plug a number on the intervals into your derivative (just like the first derivative test). *Note: If the problem says “non-negative” or something like that, start your intervals at 0. Anyway, once you plug in your numbers, decide what’s a max or min based on if you got a negative number or a positive number. Now, plug your x-values and your end points (such as 0 or, if stated, the sum) into your secondary equation and solve for the other variable. Then plug both variables into your primary equation. If you are looking for a maximum, your answer is the x-value with the biggest y-value. If you are looking for a minimum, your answer is the x-value with the smallest y-value.

Ex. Build a rectangular pen with three parallel partitions using 500 feet of fencing. What dimensions will maximize the total area of the pen?
*x = width & y = length
Primary equation: A = xy
Secondary equation: 500 = 5x + 2y
Solved secondary equation: y = 250 – (5/2)x
Final equation: A = 250x – (5/2)x²
Derivative: A’ = 250 – 5x
Solved: x = 50
Intervals: (0, 50) u (50, 100)
f’(1) = +ve
f’(51) = -ve
Plug in: If x = 0, then y = 250, and A = 0
If x = 50, then y = 125, and A = 6250
If x = 100, then y = 0, and A = 0
Result: y = 125 is the biggest
Final answer: Your maximum area is 6250 ft²

I hope everyone had a good holiday, but more Calculus awaits us...