# Thread: geometry of the derivative: applications of maximisation and minimisation

1. ## geometry of the derivative: applications of maximisation and minimisation

Here is the question: the diagram(which I am unable to draw) has two parabolas in the first quadrant, one with a minimum at (3,7) and one with a maximum at (2,4) It has a line which touches the top parabola at A and the bottom parabola at B the expression for the top parabola is y=(x-3)^2 +7 where the line joins it at A and the expression for the bottom parabola is y=x(4-x)where the line joins it at point B Point A and point B have the same x coordinate. (which means that it is perpendicular to the x axis and the expression for the line is x=? with no y value given. Find an expression for the length of AB and determine its minimum length. The length of the line AB must be an expression of the y terms which is the height of the line. A is at y=7+A and B is at y= 4-B so an expression for the length of the line is y= A+B+3 taking into account the 3y units between the maximum and the minimum of the two parabolas. What I don't understand is how to use differentiation to find the minimum length but I assume it depends on finding the value for x for the line

2. Originally Posted by Jeane
… Point A and point B have the same x coordinate … which means that [the line AB] is perpendicular to the x axis and the expression for the line is x=? with no y value given
Hi. Were you expecting some value of y to be given? Symbol y never appears in the equation for a vertical line because its value can be any Real number.

… The length of the line AB must be an expression of the y terms which is the height of the line. A is at y=7+A and B is at y= 4-B so an expression for the length of the line is y= A+B+3 taking into account the 3y units between the maximum and the minimum of the two parabolas.
Was all of this wording given to you?

If so, it seems like they want you to first find values for A and B. There is a much simpler approach (my opinion) for finding the value of x that leads to the shortest vertical distance between the parabolas.

What I don't understand is how to use differentiation to find the minimum length but I assume it depends on finding the value for x for the [vertical line passing through A and B].
I solved it by using the given quadratic functions to express the distance AB in terms of x (not in terms of A and B, which I ignored).

The expression for the distance AB (in terms of x) is a differentiable function. We can use its first derivative, to find its minimum value.

Does the given diagram look anything like this? (Vertical line omitted, for now).

The top blue line is y = A.

The bottom blue line is y = B.

The top green line is y = 7.

The bottom green line is y = 4.

jeane9.jpg

3. Originally Posted by mmm4444bot
Hi. Were you expecting some value of y to be given? Symbol y never appears in the equation for a vertical line because its value can be any Real number.

Was all of this wording given to you?

If so, it seems like they want you to first find values for A and B. There is a much simpler approach (my opinion) for finding the value of x that leads to the shortest vertical distance between the parabolas.

I solved it by using the given quadratic functions to express the distance AB in terms of x (not in terms of A and B, which I ignored).

The expression for the distance AB (in terms of x) is a differentiable function. We can use its first derivative, to find its minimum value.

Does the given diagram look anything like this? (Vertical line omitted, for now).

The top blue line is y = A.

The bottom blue line is y = B.

The top green line is y = 7.

The bottom green line is y = 4.

jeane9.jpg
[No the line is perpendicular to the x axis and joins the parabolas on the RHS of these]

4. Originally Posted by Jeane
[No the line is perpendicular to the x axis and joins the parabolas on the RHS of these]
To which of the four drawn lines are you referring? (Note that the helper specified that he was not including any vertical lines for the time being, so a line "perpendicular to the x-axis" would not be relevant at this stage.) How does this line "join" the two parabolas? Where does it do this?

Also, the helper asked you some questions, in order to clarify what may be going on; he also suggested a course of action. When you reply, please include your responses to the questions and hints. Thank you!

5. ## The minimum and maximum

thanks for your reply The figures for the minimum and maximum are given in the original drawing. Thought I gave them to you. Is there any point to finding the derivative if that is the case?

6. ## confused

[QUOTE=Jeane;423181]thanks for your reply The figures for the minimum and maximum are given in the original drawing. Thought I gave them to you. Is there any point to finding the derivative if that is the case?[/don't know how your drawing relates to the original i was working from. I already understood that the equation for the line is x=? The wording I gave for the question was mostly mine because I was trying to explain the whole situation but it is not too different from the question in the text and still asks what is the length of the shortest distance between the two parabolas]

7. ## perpendicular distance formula

Are you referring to the use of the perpendicular distance formula for the length of the line between the two parabolas?

8. Originally Posted by Jeane
No the line is perpendicular to the x axis and joins the parabolas on the RHS of these
As I stated with my graph, I did not draw the vertical line. If I had, then I would have been giving you the answer.

I took a guess at what your diagram shows; since they're talking about y=A+B+3, I considered the possibility that they drew lines for y=A and y=B.

It would be better, were you to attach a screen shot of what you're looking at.

I'm sorry, I do not understand what you're trying to express with that phrase.

9. Originally Posted by Jeane
I already understood that the equation for the line is x=?
But you inferred that y was missing, when you mentioned the equation for the vertical line. That is my point.

You said, "… the expression for the line is x=? with no y value given …"

I'm curious why you added the phrase "with no y value given". It seems to suggest that maybe you're operating under some misconception about the form of equations for vertical lines because y is never given, in the equation for a vertical line. Otherwise, there's no reason to mention anything about y.

The wording I gave for the question was mostly mine because I was trying to explain the whole situation but it is not too different from the question in the text and still asks what is the length of the shortest distance between the two parabolas
I'm sure you've read the forum guidelines, by now. We prefer that you provide the exact text of the given exercise, word for word.

I found your OP somewhat difficult to understand; I had to make guesses.

Do you know if you are free to complete this exercise any way you choose? Do you really need to first provide values for A and B? There is an easier approach, to find the distance AB.

10. Originally Posted by Jeane
Are you referring to the use of the perpendicular distance formula
I've never heard of a "perpendicular distance formula". Are you talking about the Distance Formula, which is used to find the distance between two known (x,y) points?

I'm not even sure what you're looking at in this thread, when you ask that question. Are you talking to me or stapel?

I'm not using any formula for a line. My method takes the difference of the two given Quadratic polynomials (because that expression is the vertical distance between the parabolas at each x).

I then found the first derivative of this distance expression, and I used the derivative to set up an equation to solve for x (in the usual way, using a first derivative to find a minimum). That solution for x gave me the position of the vertical line, and I substituted that value of x into the distance expression to obtain the length AB.