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## Differential Calculus

### Course: Differential Calculus>Unit 5

Lesson 11: Solving optimization problems

# Optimization problem: extreme normaline to y=x²

A difficult but interesting derivative word problem. Created by Sal Khan.

## Want to join the conversation?

• What is the source of this problem? • at how did you decide to factor out 4/x^2. It seems like a wild goose chase to me. • If you notice that a polynomial has degrees such as -2, then 0, then 2, and you want to turn it into 0, 2, and 4 to make it easier to factor, you can multiply by the second degree, but must make sure to also divide by the second degree to keep it equivalent. Dividing by the second degree is the same as "factoring out" the reciprocal of the second degree. Example, a = (1/1)a = (n/n)a = (1/n)(n)a = (1/n)(na). I don't know if this helps.
• At , why does -1 + 1/(2x_0^2) "reach a maximum or a minimum when it = 0"? • At , why is the derivative 0 at maximum point? Has Sal done videos on it? If not, can someone please explain to me why is the derivative = 0 , at minimum/maximum point? Thanks in advance. •  You can view it this way: when a function reaches a maximum point, the tangent line goes horizontally, i.e. it's slope equals 0. So, the derivative of a function at a maximum/minimum point = 0.
• What do you mean by "not" when you say "x not" Sal? •  He isn't saying "x not". He is saying "x naught". The word "naught" means zero.
The zero (or naught) is a sub-script of x, that is to say a particular fixed value of x, similar to X1, X2, X3, i.e. specific fixed values of the variable x.
• Im sure this is a very ignorant question, but what is the difference between x and x naught ? • Can someone explain please to me why did he take the derivative of the function in 18.50 I understand everything until that part. • At that point the work done has been to figure out a function that tells you what the x value at the intersection on the left (Quadrant 2) will be for a given x value at the intersection on the right (Quadrant 1). Now if you were to graph that function then you would be graphing the X naught value on the x axis and the resulting computed x value, or f(x naught) on the y axis. Since you want to find the maximum value for f(x naught) you would take a derivative of the function set to zero to show you where that curve achieves a maximum, at which point it's slope would be zero.
• at , why the equation of normal line is: y-x0^2=-(1/2x0)*(x-x0)? • Well, that is the point-slope form of the normal line,

the point being (x0, y0), and since y=x^2, y0 = x0^2, so the point is (x0, x0^2),

the slope being perpendicular (which is what "normal" means, it would be nice if Sal mentioned that) to the tangent line at (x0, y0); the slope of the tangent line is just y' = 2x, so the line perpendicular to that has the slope -1/2x (the negative reciprocal of the slope of the tangent line), which at the point x0 is m = -1/2x0

So, in summary, the equation is of the form:

y - y0 = m(x - x0)

y0 = -x0^2

m = -1/2x0
• At , Sal says x0 squared is equal to 1/2.
Surely this impies that x0 is equal to plus OR minus 1/sqrt2?
Obviously that would put it in the wrong quadrant, but is there a way to mathematically justify discarding a solution like this?  