1) You must find the derivative because you are looking for the maximum or minimum. Sometimes this requires using two equations if you have more than one variable. You may have to use information in the problem to create two equations and then determine which one is used to find the derivative and which one is used to find a variable. For example, if you are finding a point that is closest, you are finding the minimum and will use the distance formula to find the derivative.
2) Once you find the derivative, you must set it equal to 0.
3) Next, you solve to find the minimum or maximum. Sometimes you will have more than one possible answer and must determine which is correct based on information in the problem.
Here is an Example
Find the point on the line y = 2x + 3 that is closest to the origin (0,0).
1) The first step is to find the derivative. Due to the fact that we are finding the point closest to the origin, the distance formula is one of our equations.
d = √(x2 - x1)2 + (y2 - y1)2
d = √(x2 - 0)2 + (2x + 3 - 0)2
Now we will square both sides in order to remove the square root.
d2 = (x)2 + (2x + 3)2
Now we will find the derivative using the power rule and the chain rule.
2d = 2x + 2(2x + 3)(2)
2d = 2x + 4(2x + 3)
2d = 2x + 8x + 12
2d = 10x + 12
Now we will set the derivative equal to 0 and will solve for x.
0 = 10x + 12
-12 = 10x
x = -12/10 = -6/5
Now we will plug in this x-value in order to find the y-value of the point.
y = 2(-6/5) + 3
y = 3/5
This means that the closest point on the line y = 2x + 3 to the origin is (-6/5, 3/5).
If f'(x) = 6x2 - 10x - 1 give two options of what f(x) could be
f(x) = 2x3 - 5x2 - x
f(x) = 2x3 - 5x2 - x + 1000
I like how you included a detailed explanation both separate from the example and with the example. You cover every detail of the optimization process. Good work Sweese.
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