# how to maximize the value of z?

Discussion in 'Homework Help' started by PG1995, Oct 29, 2017.

1. ### PG1995 Thread Starter Distinguished Member

Apr 15, 2011
813
6
Hi,

When a distance of 1 unit is moved along x-axis, there is 1 unit height increase along z-axis. Likewise, when a distance of 1 unit is moved along y-axis, there is 2 unit height increase along z-axis. It means that when you move a distance of √2 along diagonal, there would be a 3-unit height increase along z-axis. Please see the the picture.

We only want to move 1-unit along the diagonal and still want the height along z-axis to be maximum. It means that we need a certain combination of x- and y-units.

I think that 0.45 units along x-axis and 0.9 units along y-axis will give us the maximum height along z which is 2.25. Please have a look here.

I'm not sure if the combination given above is correct, and also how the method works. It looks like an optimization problem. Not sure. Could you please guide me? Thank you.

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2. ### MrAl AAC Fanatic!

Jun 17, 2014
6,383
1,382

Hi there,

There are different ways to handle this. I think 'maybe' the first way to approach this is using an idea that is known simply as constrained min and max. I say this because the method is self contained and analytical unlike some methods that use steps and trial and errors to get to the solution. Those methods are good too though when the problem is not as well defined in a more analytical way.

The first requirement for this or any other method to be successful though is that there must be a min or max that actually exists. A perfectly linear function such as y=3*x will not have a min or max however a parabolic curve such as y=3*x^2 will have one.
The reason i am bringing this up is because your example seems to be somewhat undefined. On the one hand it looks like a box, on the other hand it looks like a linear function along the diagonal, so maybe a little clarifcation would help. In other words, WHAT exactly are you trying to optimize here?

I can guess that you might be trying to find the least surface area for a given volume or something like that, but this is a thing that needs to be well defined first so we know what we are looking for.

A simple example as above is a parabolic curve (two dimensions with just one independent variable) where we try to find a min or max. But once you define what you are trying to optimize we will have a better idea what to do here.

Since z is the dependent variable then z must reach some min or max when x and y are varied over the allowed range. The top half of a sphere would be a simple example where we want to find the value of x and value of y that gives us the maximum height. Similarly, the bottom half would show a min for some value of x and y.
There could be a global min or max as well as several local min or max's.

Last edited: Oct 29, 2017
3. ### WBahn Moderator

Mar 31, 2012
24,555
7,691
On what basis do you claim that moving one unit in the x and the y produces a change in height of 3? You seem to be saying that z = x + 2y, but on what basis are you concluding that. There are LOTS of functions that would give you the states increase along just the x and y axes but do anything off axis.

You say that you want to move along the diagonal -- that constrains you to moving the same distance in the x and in the y, otherwise you aren't on the diagonal.

PG1995 likes this.
4. ### wayneh Expert

Sep 9, 2010
16,102
6,219
Is the first figure part of the original question? I would read the question with the figure as setting up a simple geometry-of-triangles question. The triangle has rise of 3 and a run of √2. You are asked for the rise (and run) that would take you just 1 unit along the hypotenuse. The word "maximum" is confusing but could just mean "riding along the hypotenuse".

I believe your answer is incorrect because, at any point on the hypotenuse, y = x, and z = 3x.

The mile-high view of this problem is defining what is asked for.

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5. ### PG1995 Thread Starter Distinguished Member

Apr 15, 2011
813
6
Thank you, everyone.

I just made it up so I think that this is the best I could do. I might have read something which gave rise to this question.

When a distance of 1 unit is moved along x-axis, there is 1 unit height increase along z-axis. Likewise, when a distance of 1 unit is moved along y-axis, there is 2 unit height increase along z-axis. It means that when you move a distance of √2 along diagonal, there would be a 3-unit height increase along z-axis. Please see the the picture.

We only want to move 1-unit along some diagonal distance and still want the height along z-axis to be maximum. It means that we need a certain combination of x- and y-units.

I think that 0.45 units along x-axis (1-z/1-x x 0.45-x=0.45-z) and 0.9 units along y-axis (2-z/1-y x 0.9-y=1.8-z) will give us the maximum height along z which is 2.25. Please have a look here.

In this case somehow rise/run ratio for the x and y directions determine their scaling for a unit distance in a diagonal direction.

Last edited: Oct 29, 2017
6. ### MrAl AAC Fanatic!

Jun 17, 2014
6,383
1,382
Hi,

Geeze, you are going to have to clarify all of this or else you'll get nowhere.

What the heck is:
(1-z/1-x x 0.45-x=0.45-z)

That's not an equation, so what the heck is it?

Also, there is no min or max for a perfectly linear function. If you are talking delta x and delta y then you are really talking about the an actual function so it would be good to look at or a representative function like z=3*x*y+1 or something.

Is this what you are dealing with perhaps...
If you move 1 unit along x and get 1, and move 2 units along y and get 2, then:
pz/px=1
pz/py=2

where pz/px is partial derivative of z with respect to x, and ditto for pz/py with y.

If you clarify what you are trying to do we can solve this pretty quick.
Are you trying to find the unit vector by any chance?

7. ### PG1995 Thread Starter Distinguished Member

Apr 15, 2011
813
6
The following is the solution to the problem stated in post #1.

When a distance of 1 unit is moved along x-axis, there is 1 unit height increase along z-axis. Likewise, when a distance of 1 unit is moved along y-axis, there is 2 unit height increase along z-axis.

Given the information above:
height along z = x-distance + 2*y-distance, or, z=x+2y

Pythagoras theorem:
hypotenuse^2 = base^2 + perpendicular^2
diagonal-distance^2 = x-distance^2 + y-distance^2
y-distance^2 = diagonal^2 - x-distance^2
y-distance = sqrt{1 - x-distance^2}

therefore,
z=x+2√(1 - x^2}

Now using calculus.

y-distance = sqrt{1 - x-distance^2}=sqrt{1 - 0.45^2}=0.9

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