Hockey Rink Optimization problem

[CourtneyS]

Homework Statement

[/B]
Solve optimization problem;
A person is to skate up a hockey rink, in a straight line(parallel to both of the sides), from wherever they are standing. They are to shoot the puck into the net whenever they reach the point where the angle on the net is the largest.
Need to find how far up the ice the person should skate to get the largest possible angle on the net.

Homework Equations

To get the largest angle I drew a triangle from the goal line to some point where the person would stand.
Then I drew a line down the middle of the triangle to create 2 right triangles.
Length of goal line is 1.82 m, half of that is 0.91 m.
SO.. I got that sin(theta/2)=0.91/hypotenuse and cos(theta/2)=height/hypotenuse
We have that AREA=1/2 base*height
SO, if we sovle for height we get height=hypotenuse*cos(theta/2) .. solve for hypotenuse we get hypotenuse=0.91/sin(theta/2)
Plug this into area formula and we get
A=1/2*0.91*(0.91/sin(theta/2))*cos(theta/2)
simplify
A=0.41405*cot(theta/2)
SO if I wanted the max angle it would be between 0 and pi so to get max angle I would take derivative of A
which is
A'=0.207025*(-csc^2(theta/2))
So max/min values occur when A'=0 (yes), but do they also occur when A'=undefined because A' not equal to 0 ever (as far as I know) .
So I guess I'm wondering if I'm on the right track..
And then I guess once I have the angle all I have to do is plug it in and get distance.
I really just want to know if I'm on the right track cause this doesn't really seem all that much like an optimization question.. I guess it is ... thanks

The Attempt at a Solution

Oops attempt is up there^^^ sorry

 

[RUber]

I am having trouble visualizing what you did. I assume you will be given an x on the rink maybe x in [-a,a] and you have to find y>0, placing zero at the goal line, where the angle is maximized?
I imagine you should be able to write a formula for the angle between lines from a point (x,y) to (-.91,0) and (.91,0). Holding x constant, you would optimize angle by varying y.
It looks to me that if x were anywhere in line with the goal line itself, y=0 would give the max with theta = pi.

 

[CourtneyS]

I am having trouble visualizing what you did. I assume you will be given an x on the rink maybe x in [-a,a] and you have to find y>0, placing zero at the goal line, where the angle is maximized?
I imagine you should be able to write a formula for the angle between lines from a point (x,y) to (-.91,0) and (.91,0). Holding x constant, you would optimize angle by varying y.
It looks to me that if x were anywhere in line with the goal line itself, y=0 would give the max with theta = pi.

This is what I did by drawing the triangle and the top of it is where the person should stand when they shoot, and the base of the triangle is the length of the goal line 1.82 m. Then divided it in half to make 2 right triangles and theta/2 is labelled on each side.

 

[RUber]

In that case, it looks like your work above is to maximize the area, not the angle.
If you start in front of the goal, and move toward the goal, your maximum angle is $\pi$ when you reach the goal line. Your maximum area calculation would lead you to a smaller $\theta$ and a larger distance from the goal.
Reading the problem, it seems like the initial point might be anywhere on the ice, so you are looking to maximize angle for a fixed x by changing y.
If I am not misreading the problem, try setting it up as I described above, for any given x on the ice, setting your origin at the center of the goal.

 

[Ray Vickson]

Homework Statement

[/B]
Solve optimization problem;
A person is to skate up a hockey rink, in a straight line(parallel to both of the sides), from wherever they are standing. They are to shoot the puck into the net whenever they reach the point where the angle on the net is the largest.
Need to find how far up the ice the person should skate to get the largest possible angle on the net.

Homework Equations

To get the largest angle I drew a triangle from the goal line to some point where the person would stand.
Then I drew a line down the middle of the triangle to create 2 right triangles.
Length of goal line is 1.82 m, half of that is 0.91 m.
SO.. I got that sin(theta/2)=0.91/hypotenuse and cos(theta/2)=height/hypotenuse
We have that AREA=1/2 base*height
SO, if we sovle for height we get height=hypotenuse*cos(theta/2) .. solve for hypotenuse we get hypotenuse=0.91/sin(theta/2)
Plug this into area formula and we get
A=1/2*0.91*(0.91/sin(theta/2))*cos(theta/2)
simplify
A=0.41405*cot(theta/2)
SO if I wanted the max angle it would be between 0 and pi so to get max angle I would take derivative of A
which is
A'=0.207025*(-csc^2(theta/2))
So max/min values occur when A'=0 (yes), but do they also occur when A'=undefined because A' not equal to 0 ever (as far as I know) .
So I guess I'm wondering if I'm on the right track..
And then I guess once I have the angle all I have to do is plug it in and get distance.
I really just want to know if I'm on the right track cause this doesn't really seem all that much like an optimization question.. I guess it is ... thanks

The Attempt at a Solution

Oops attempt is up there^^^ sorry

Suppose the goal is along the x-axis and its ends are at (-a,0) and (+a,0). Suppose the skater starts at (b,Y) (Y > 0 large) and skates down the line x = b to the point (b,y), where he then shoots the puck. Assume b > 0, so the skater is to the right of the center line of the goal.

Obviously, if b < a the skater should just skate down to y = 0, because that will make the "goal angle" equal to 180 degrees. If b = a he should skate down to y = 0 because that makes the "goal angle" equal to 90 degrees. So, assume that b > a. Now you have a non-trivial problem, but it is not that difficult: goal angle = angle to right-end of goal - angle to left-end of goal, and each of those angles is easy enough to get, just by looking at the geometry.

 

[CourtneyS]

In that case, it looks like your work above is to maximize the area, not the angle.
If you start in front of the goal, and move toward the goal, your maximum angle is $\pi$ when you reach the goal line. Your maximum area calculation would lead you to a smaller $\theta$ and a larger distance from the goal.
Reading the problem, it seems like the initial point might be anywhere on the ice, so you are looking to maximize angle for a fixed x by changing y.
If I am not misreading the problem, try setting it up as I described above, for any given x on the ice, setting your origin at the center of the goal.

Sorry, my first attempt at the solution was wrong. I am definitely looking to maximize theta.
I have done something different now that I think will help me maximize theta.
Attached is a diagram of what I am trying to achieve..
The triangle does not have to be a right triangle since it will occur at whatever x and y coordinate that will give max theta.
So I have drawn a right triangle and and labbeled the sides on the new triangle x and y to try and achieve a solution.
Here is my start at the solution.
tan(alpha) = y/x
alpha = arctan(y/x)

tan(theta+alpha)=(1.82+y)/x
alpha + theta = arctan((1.82+y)/x)

theta=arctan((1.82+y)/2) - arctan(y/x)

Looking to get theta' (so take derivative) and then set =0 and solve for the parameters when the equation = 0 to find max value.
I am unsure whether y is a constant or not?
I don't think it is so I would have to treat both as variables and take derivative of fx and fy and find out what makes the partials = 0.

I am sort of following an example from this webpage but they have the height of their triangle so it is a bit different/easier than my problem.
https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/maxminsol2directory/MaxMinSol2.html

It it labeled SOLUTION 14.

Any help you can provide will be great thanks

 

[Ray Vickson]

Sorry, my first attempt at the solution was wrong. I am definitely looking to maximize theta.
I have done something different now that I think will help me maximize theta.
Attached is a diagram of what I am trying to achieve..
The triangle does not have to be a right triangle since it will occur at whatever x and y coordinate that will give max theta.
So I have drawn a right triangle and and labbeled the sides on the new triangle x and y to try and achieve a solution.
Here is my start at the solution.
tan(alpha) = y/x
alpha = arctan(y/x)

tan(theta+alpha)=(1.82+y)/x
alpha + theta = arctan((1.82+y)/x)

theta=arctan((1.82+y)/2) - arctan(y/x)

Looking to get theta' (so take derivative) and then set =0 and solve for the parameters when the equation = 0 to find max value.
I am unsure whether y is a constant or not?
I don't think it is so I would have to treat both as variables and take derivative of fx and fy and find out what makes the partials = 0.

I am sort of following an example from this webpage but they have the height of their triangle so it is a bit different/easier than my problem.
https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/maxminsol2directory/MaxMinSol2.html

It it labeled SOLUTION 14.

Any help you can provide will be great thanks

Going by the notation in your diagram, y is constant, and x is variable, so $f_y$ is meaningless. Only $f_x$ has any significance in this problem.

 

[CourtneyS]

Going by the notation in your diagram, y is constant, and x is variable, so $f_y$ is meaningless. Only $f_x$ has any significance in this problem.

Ok, just to conf

Going by the notation in your diagram, y is constant, and x is variable, so $f_y$ is meaningless. Only $f_x$ has any significance in this problem.

Ok, just to confirm:
The person can start off at any point horizontally or vertically, but once they start moving they can only move vertically, and then they stop at the point where theta is maximized.
So, y is constant if this is the case, y being labelled in the diagram?

And I can just take the derivative treating y as a constant and that is how I will find my max theta?

 

[RUber]

Yes.
If both x and y were variable, you would always end up right in front of the net--angle of 180 degrees.

 

[CourtneyS]

Yes.
If both x and y were variable, you would always end up right in front of the net--angle of 180 degrees.

The goal crease has a radius of 1.8m and goes around the front of the net like a semi circle, the person can only shoot from 1.8m away if they skate forward in a straight line. So the 180deg thing cannot occur for this case.

 

[Ray Vickson]

The goal crease has a radius of 1.8m and goes around the front of the net like a semi circle, the person can only shoot from 1.8m away if they skate forward in a straight line. So the 180deg thing cannot occur for this case.

My interpretation of the problem was a bit different: I assumed the skater starts at some fixed, specified distance from the sides of the rink, but can skate only parallel to the sides.

You seem to be allowing the skater to also choose the distance from the rink sides as another variable. In that case, the best distance is half-way across (i.e., on the line through the center of the net), but the best distance from the end of the rink is right at the semicircle around the net's center. In that case the derivative (in the direction of the sides) is NOT equal to zero, because you have an end-point maximum. (However, for any fixed distance from the end, the derivative along the direction of the ends IS zero).