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## Differential Calculus

### Course: Differential Calculus > Unit 5

Lesson 11: Solving optimization problems- Optimization: sum of squares
- Optimization: box volume (Part 1)
- Optimization: box volume (Part 2)
- Optimization: profit
- Optimization: cost of materials
- Optimization: area of triangle & square (Part 1)
- Optimization: area of triangle & square (Part 2)
- Optimization problem: extreme normaline to y=x²
- Optimization
- Motion problems: finding the maximum acceleration

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# 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?(29 votes)
- And we've got solve it in under 3-5 mins for 6 marks , although(9 votes)

- at13:20how did you decide to factor out 4/x^2. It seems like a wild goose chase to me.(29 votes)
- 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.(20 votes)

- At20:11, why does -1 + 1/(2x_0^2) "reach a maximum or a minimum when it = 0"?(30 votes)
- When the derivative of a function is zero, that means that the slope of the function is zero, so it has reached a local maximum or minimum value. As seen here: http://imgur.com/vw9uW(21 votes)

- At18:45, 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.(14 votes)
- 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.(28 votes)

- What do you mean by "not" when you say "x not" Sal?(9 votes)
- 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.(37 votes)

- Im sure this is a very ignorant question, but what is the difference between x and x naught ?(9 votes)
- x refers to the variable x in general. x-naught refers to a specific value of x. We don't know what that specific value is, so we can't say 3 or 4 or 1.57. So we call it x-naught.(4 votes)

- Can someone explain please to me why did he take the derivative of the function in 18.50 I understand everything until that part.

Thanks in advance.(6 votes)- 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.(3 votes)

- at08:25, why the equation of normal line is: y-x0^2=-(1/2x0)*(x-x0)?(6 votes)
- 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(2 votes)

- At21:06, 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?(4 votes)- You just answered your own question. x0 is defined as the interception point in the first quadrant, thus is has to be positive.If you are on your exams and feel like clarifying just in case, you could add in an extra line, ie."Because x0 is defined to be in the first quadrant, x0 > 0. Therefore, x0 = 1/sqrt2"

Hope that helps(4 votes)

- Aren't there quicker ways to solve this problem? For instance, if we are looking for the roots of x² + (1/2x₀) x - (1/2 + x₀²), we already know that x = x₀ is one of the roots because the normal line obviously intersects the parabola at (x₀, x₀²). So what's the other root? Well, from the quadratic equation we know either that the sum of the two roots is -1/2x₀ or that the product of the two roots is -(1/2 + x₀²), and either of those get us to the other root at -x₀ - 1/(2x₀) without needing to use the Quadratic Formula at all.(5 votes)
- Yes, theoretically you could, but do you really want to try to factor that incredibly hairy equation based on guess and check factoring? Overall, the quadratic formula is more surefire and probably will actually take less time. If this problem was on a timed test and I was aware of both methods, I would choose the quadratic formula, just so that I don't have to guess at all. I'm assuming that you saw this method after you learned what the other root was.(2 votes)

## Video transcript

I just got sent this
problem, and it's a pretty meaty problem. A lot harder than what
you'd normally find in most textbooks. So I thought it would help
us all to work it out. And it's one of those problems
that when you first read it, your eyes kind of glaze over,
but when you understand what they're talking about, it's
reasonably interesting. So they say, the curve in the
figure above is the parabola y is equal to x squared. So this curve right there is
y is equal to x squared. Let us define a normal line as
a line whose first quadrant intersection with the parabola
is perpendicular to the parabola. So this is the first
quadrant, right here. And they're saying that a
normal line is something, when the first quadrant intersection
with the parabola is normal to the parabola. So if I were to draw a tangent
line right there, this line is normal to that tangent line. That's all that's saying. So this is a normal
line, right there. Normal line. Fair enough. 5 normal lines are
shown in the figure. 1, 2, 3, 4, 5. Good enough. And these all look
perpendicular, or normal to the parabola in the first
quadrant intersection, so that makes sense. For a while, the x-coordinate
of the second quadrant intersection of the normal line
of the parabola gets smaller, as the x-coordinate of the
first quadrant intersection gets smaller. So let's see what happens as
the x-quadrant of the first intersection gets smaller. So this is where I left
off in that dense text. So if I start at this point
right here, my x-coordinate right there would look
something like this. Let me go down. my x-coordinate is
right around there. And then as I move to a smaller
x-coordinate to, say, this one right here, what happened
to the normal line? Or even more important, what
happened to the intersection of the normal line in
the second quadrant? This is the second
quadrant, right here. So when I had a larger
x-value here, my normal line intersected here,
in the second quadrant. Then when I brought my x-value
in, when I lowered my x-value, my x-value here, because this
is the next point, right here, my x-value at the intersection
here, went-- actually, their wording is bad. They're saying that
the second quadrant intersection gets smaller. But actually, it's not
really getting smaller. It's getting less negative. I guess smaller could be just
absolute value or magnitude, but it's just getting
less negative. It's moving there, but
it's actually becoming a larger number, right? It's becoming less negative,
but a larger number. But if we think in absolute
value, I guess it's getting smaller, right? As we went from that point to
that point, as we moved the x in for the intersection of the
first quadrant, the second quadrant intersection also
moved in a bit, from that line to that line. Fair enough. But eventually, a normal line
second quadrant intersection gets as small as it can get. So if we keep lowering our
x-value in the first quadrant, so we keep on pulling in the
first quadrant, as we get to this point. And then this point intersects
the second quadrant, right there. And then, if you go even
smaller x-values in the first quadrant then your normal line
starts intersecting in the second quadrant, further and
further negative numbers. So you can kind of view this
as the highest value, or the smallest absolute value, at
which the normal line can intersect in the
second quadrant Let me make that clear. Up here, you were intersecting
when you had a large x in the first quadrant, you had a large
negative x in the second quadrant intersection. And then as you lowered your
x-value, here, you had a smaller negative value. Up until you got to this point,
right here, you got this, which you can view as the smallest
negative value could get, and then when you pulled in your x
even more, these normal lines started to push out again,
out in the second quadrant. That's, I think, what
they're talking about. The extreme normal line
is shown as a thick line in the figure. Right. This is the extreme normal
line, right there. So this is the extreme
one, that deep, bold one. Extreme normal line. After this point, when you pull
in your x-values even more, the intersection in your second
quadrant starts to push out some. And you can think of the
extreme case, if you draw the normal line down here, your
intersection with the second quadrant is going to be way out
here someplace, although it seems like it's kind of
asymptoting a little bit. But I don't know. Let's read the rest
of the problem. Once the normal line passes
the extreme normal line, the x-coordinates of their second
quadrant intersections what the parabola start to increase. And they're really, when they
say they start to increase, they're actually just
becoming more negative. That wording is bad. I should change this to
more, more negative. Or they're becoming
larger negative numbers. Because once you get below
this, then all of a sudden the x-intersections start
to push out more in the second quadrant. Fair enough. The figures show 2
pairs of normal lines. Fair enough. The 2 normal lines of a pair
have the same second quadrant intersection with the parabola,
but 1 is above the extreme normal line, in the first
quadrant, the other is below it. Right, fair enough. For example, this guy right
here, this is when we had a large x-value. He intersects with the
second quadrant there. Then if you lower and lower the
x-value, if you lower it enough, you pass the extreme
normal line, and then you get to this point, and then this
point, he intersects, or actually, you go to this point. So if you pull in your x-value
enough, you once again intersect at that same point
in the second quadrant. So hopefully I'm making some
sense to you, as I try to make some sense of this problem. OK. Now what do they want to know? And I think I only have time
for the first part of this. Maybe I'll do the second
part in the another video. Find the equation of the
extreme normal line. Well, that seems very daunting
at first, but I think our toolkit of derivatives, and
what we know about equations of a line, should be
able to get us there. So what's the slope of
the tangent line at any point on this curve? Well, we just take the
derivative of y equals x squared, and y prime
is just equal to 2x. This is the slope of the
tangent at any point x. So if I want to know the slope
of the tangent at x0, at some particular x, I would just
say, well, let me just say, slope, it would be 2 x0. Or let me just say, f of
x0 is equal to 2 x0. This is the slope at
any particular x0 of the tangent line. Now, the normal line slope
is perpendicular to this. So the perpendicular line, and
I won't review it here, but the perpendicular line has a
negative inverse slope. So the slope of normal line at
x0 will be the negative inverse of this, because this is the
slope of the tangent line x0. So it'll be equal to
minus 1 over 2 x0. Fair enough. Now, what is the equation of
the normal line at x0 let's say that this is my x0 in question. What is the equation of
the normal line there? Well, we can just use
the point-slope form of our equation. So this point right here
will be on the normal line. And that's the
point x0 squared. Because this the graph of
y equals x0, x squared. So this normal line will
also have this point. So we could say that the
equation of the normal line, let me write it down, would
be equal to, this is just a point-slope definition
of a line. You say, y minus the y-point,
which is just x0 squared, that's that right there, is
equal to the slope of the normal line minus 1 over
2 x0 times x minus the x-point that we're at. Minus x, minus x0. This is the equation
of the normal line. So let's see. And what we care about is when
x0 is greater than 0, right? We care about the normal line
when we're in the first quadrant, we're in all of
these values right there. So that's my equation
of the normal line. And let's solve it
explicitly in terms of x. So y is a function of x. Well, if I add x0 squared
to both sides, I get y is equal to, actually, let
me multiply this guy out. I get minus 1/2 x0 times x, and
then I have plus, plus, because I have a minus times
a minus, plus 1/2. The x0 and the over the
x0, they cancel out. And then I have to add
this x0 to both sides. So all I did so far, this is
just this part right there. That's this right there. And then I have to add this to
both sides of the equation, so then I have plus x0 squared. So this is the equation of the
normal line, in mx plus b form. This is its slope, this is
the m, and then this is its y-intercept right here. That's kind of the b. Now, what do we care about? We care about where
this thing intersects. We care about where it
intersects the parabola. And the parabola, that's pretty
straightforward, that's just y is equal to x squared. So to figure out where they
intersect, we just have to set the 2 y's to be
equal to each other. So they intersect, the x-values
where they intersect, x squared, this y would have
to be equal to that y. Or we could just substitute
this in for that y. So you get x squared is equal
to minus 1 over 2 x0 times x, plus 1/2 plus x0 squared. Fair enough. And let's put this in a
quadratic equation, or try to solve this, so we can apply
the quadratic equation. So let's put all of this
stuff on the lefthand side. So you get x squared plus
1 over 2 x0 times x minus all of this, 1/2 plus x0
squared is equal to 0. All I did is, I took all of
this stuff and I put it on the lefthand side of the equation. Now, this is just a standard
quadratic equation, so we can figure out now where the
x-values that satisfy this quadratic equation will tell
us where our normal line and our parabola intersect. So let's just apply the
quadratic equation here. So the potential x-values,
where they intersect, x is equal to minus b, I'm
just applying the quadratic equation. So minus b is minus 1 over
2 x0, plus or minus the square root of b squared. So that's that squared. So it's one over four
x0 squared minus 4ac. So minus 4 times 1 times
this minus thing. So I'm going to have a minus
times a minus is a plus, so it's just 4 times this,
because there was one there. So plus 4 times
this, right here. 4 times this is just
2 plus 4 x0 squared. All I did is, this
is 4ac right here. Well, minus 4ac. The minus and the minus
canceled out, so you got a plus. There's a 1. So 4 times c is just
2 plus 4x squared. I just multiply this by two,
and of course all of this should be over 2 times
a. a is just 2 there. So let's see if I
can simplify this. Remember what we're doing. We're just figuring out where
the normal line and the parabola intersect. Now, what do we get here. This looks like a little
hairy beast here. Let me see if I can simplify
this a little bit. So let us factor out--
let me rewrite this. I can just divide everything by
1/2, so this is minus 1 over 4 x0, I just divided this by 2,
plus or minus 1/2, that's just this 1/2 right there, times the
square root, let me see what I can simplify out of here. So if I factor out a 4 over
x0 squared, then what does my expression become? This term right here will
become an x to the fourth, x0 to the fourth, plus, now,
what does this term become? This term becomes
a 1/2 x0 squared. And just to verify this,
multiply 4 times 1/2, you get 2, and then the x0
squares cancel out. So write this term times that,
will equal 2, and then you have plus-- now we factored a four
out of this and the x0 squared, so plus 1/16. Let me scroll over
a little bit. And you can verify
that this works out. If you were to multiply this
out, you should get this business right here. I see the home stretch here,
because this should actually factor out quite neatly. So what does this equal? So the intersection of
our normal line and our parabola is equal to this. Minus 1 over 4 x0 plus or
minus 1/2 times the square root of this business. And the square root, this
thing right here is 4 over x0 squared. Now what's this? This is actually, lucky
for us, a perfect square. And I won't go into details,
because then the video will get too long, but I think you can
recognize that this is x0 squared, plus 1/4. If you don't believe me,
square this thing right here. You'll get this
expression right there. And luckily enough, this is
a perfect square, so we can actually take the
square root of it. And so we get, the point
at which they intersect, our normal line and our
parabola, and this is quite a hairy problem. The points where they intersect
is minus 1 over 4 x0, plus or minus 1/2 times the
square root of this. The square root of this is the
square root of this, which is just 2 over x0 times the square
root of this, which is x0 squared plus 1/4. And if I were to rewrite all of
this, I'd get minus 1 over 4 x0 plus, let's see, this 1/2 and
this 2 cancel out, right? So these cancel out. So plus or minus, now I
just have a one over x0 times x0 squared. So I have 1 over x0-- oh sorry,
let me, we have to be very careful there-- x0 squared
divided by x0 is just x0, let me do that in a yellow color so
you know what I'm dealing with. This term multiplied by this
term is just x0, and then you have a plus 1/4 x0. And this is all a
parentheses here. So these are the two points at
which the normal curve and our parabola intersect. Let me just be very clear. Those 2 points are, for if
this is my x0 that we're dealing with, right there. It's this point and this point. And we have a plus or minus
here, so this is going to be the plus version, and this is
going to be the minus version. In fact, the plus version
should simplify into x0. Let's see if that's the case. Let's see if the plus version
actually simplifies to x0. So these are our two points. If I take the plus version,
that should be our first quadrant intersection. So x is equal to minus 1/4
x0 plus x0 plus 1/4 x0. And, good enough, it does
actually cancel out. That cancels out. So x0 is one of the points
of intersection, which makes complete sense. Because that's how we even
defined the problem. But, so this is the first
quadrant intersection. So that's the first
quadrant intersection. The second quadrant
intersection will be where we take the minus
sign right there. So x, I'll just call it in the
second quadrant intersection, it'd be equal to minus 1/4 x0
minus this stuff over here, minus the stuff there. So minus x0 minus
1 over 4 mine x0. Now what do we have? So let's see. We have a minus 1 over 4
x0, minus 1 over 4 x0. So this is equal to minus x0,
minus x0, minus 1 over 2 x0. So if I take minus 1/4 minus
1/4, I get minus 1/2. And so my second quadrant
intersection, all this work I did got me this result. My second quadrant
intersection, I hope I don't run out of space. My second quadrant
intersection, of the normal line and the parabola, is
minus x0 minus 1 over 2 x0. Now this by itself is a pretty
neat result we just got, but we're unfortunately not
done with the problem. Because the problem wants us to
find that point, the maximum point of intersection. They call this the
extreme normal line. The extreme normal line is
when our second quadrant intersection essentially
achieves a maximum point. I know they call it the
smallest point, but it's the smallest negative value, so
it's really a maximum point. So how do we figure out
that maximum point? Well, we have our second
quadrant intersection as a function of our
first quadrant x. I could rewrite this as, my
second quadrant intersection as a function of x0 is equal to
minus x minus 1 over 2 x0. So this is going to reach a
minimum or a maximum point when its derivative is equal to 0. This is a very unconventional
notation, and that's probably the hardest
thing about this problem. But let's take this derivative
with respect to x0. So my second quadrant
intersection, the derivative of that with respect to x0, is
equal to, this is pretty straightforward. It's equal to minus 1, and then
I have a minus 1/2 times, this is the same thing as
x to the minus 1. So it's minus 1 times x0
to the minus 2, right? I could have rewritten
this as minus 1/2 times x0 to the minus 1. So you just put its
exponent out front and decrement it by 1. And so this is the derivative
with respect to my first quadrant intersection. So let me simplify this. So x, my second quadrant
intersection, the derivative of it with respect to my first
quadrant intersection, is equal to minus 1, the minus 1/2 and
the minus 1 become a positive when you multiply them, and so
plus 1/2 over x0 squared. Now, this'll reach a maximum
or minimum when it equals 0. So let's set that equal to
0, and then solve this problem right there. Well, we add one to both sides. We get 1 over 2 x0 squared is
equal to 1, or you could just say that that means that 2 x0
squared must be equal to 1, if we just invert both
sides of this equation. Or we could say that x0 squared
is equal to 1/2, or if we take the square roots of both sides
of that equation, we get x0 is equal to 1 over the
square root of 2. So we're really, really,
really close now. We've just figured out the
x0 value that gives us our extreme normal line. This value right here. Let me do it in a
nice deeper color. This value right here, that
gives us the extreme normal line, that over there is
x0 is equal to 1 over the square root of 2. Now, they want us to figure
out the equation of the extreme normal line. Well, the equation of the
extreme normal line we already figured out right here. It's this. The equation of the normal line
is that thing, right there. So if we want the equation of
the normal line at this extreme point, right here, the one that
creates the extreme normal line, I just substitute 1 over
the square root of 2 in for x0. So what do I get? I get, and this is the home
stretch, and this is quite a beast of a problem. y minus x0 squared. x0 squared is 1/2, right? 1 over the square root
of 2 squared is 1/2. Is equal to minus 1 over 2 x0. So let's be careful here. So minus 1/2 times 1 over x0. One over x0 is the square
root of 2, right? All of that times x minus x0. So that's 1 over the square
root of 2. x0 is one over square root of 2. So let's simplify
this a little bit. So the equation of our normal
line, assuming I haven't made any careless mistakes, is
equal to, so y minus 1/2 is equal to, let's see. If we multiply this minus
square root of 2 over 2x, and then if I multiply these
square root of 2 over this, it becomes one. And then I have a minus
and a minus, so that I have a plus 1/2. I think that's right. Yeah, plus 1/2, this
times this times that is equal to plus 1/2. And then, we're at
the home stretch. So we just add 1/2 to both
sides of this equation, and we get our extreme normal line
equation, which is y is equal to minus square
root of 2 over 2x. If you add 1/2 to both sides of
this equation, you get plus 1. And there you go. That's the equation of that
line there, assuming I haven't made any careless mistakes. But even if I have, I think you
get the idea of hopefully how to do this problem, which
is quite a beastly one.