Oh, I've got work for the other as well.
The center of the circular table is the origin. The equation for the edge of the table is x^2 + y^2 = r^2. Therefore, the equations for the walls it is flush to can be described as x = r and y = r.

sqrt((x - r)^2 + (r - r)^2) = 5; i.e., sqrt((x - r)^2) = 5.
So, similarly, sqrt((y - r)^2) = 10.

sqrt(a^2) = a or -a. Thus:

x - r = 5 or -5 and
y - r = 10 or -10, so

x = r + 5 or x = r - 5 and
y = r + 10 and y = r - 10.

This gives us four possibilities to look at. (Both addition, both subtraction, or a mix, in either order.) Substitute into the formula for the edge of the table.

For example, solve (r + 5)^2 + (r + 10)^2 = r^2.

Now, it's not the algebraic work that is necessarily the issue. The set-up is as important, if not more so. So, I am not going to demonstrate here that I can solve quadratic equations.

We wind up with the following possibilities, most of which are rejected, since r > 0.

(1) -5, -25
(2) 5 + 10 * I, 5 - 10 * I
(3) 25, 5
(4) -5 + 10 * I, -5 - 10 * I

Obviously, it is case 3, coming from (r - 5)^2 + (r - 10)^2 = r^2, that yields our two (viable) solutions.

This is one way to show that there really are only two solutions. When someone posing a problem makes such a claim, it should be verified.
On the other hand, besides visualization (which allowed me to also see the first solution posted was a correct one), what other approaches could there be to this problem? Algebra is a powerful tool, but solving this problem this way does not give me a feeling of satisfaction in elegance.

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Capitalize on this good fortune, one word can bring you round ... changes.

Sorry Sarge, I posted those two questions while I was sleep deprived, not really the best way to go about it, but oh well. Luckily I've had some sleep now....

Solutions and working would be desired on questions, please.

Well done to Bill, for answering Q40 correctly! (1 Point)
Q40: 40 hours.

Well done to Guido.man!, for answering the other answer for Q39 correctly. (1 Point)
Q39 Answer 2: 5 inches

A big congratulations for Sarge!, for taking the time to do the working for Question 40 Solution. (1/2 a point)

And 'ere ar' the standins':
1. William Rothamel - 9 1/2 Points
2. Mr. Kevvy - 7 Points
3. Guido.Man - 6 Points
4. Sarge - 5 1/2 Points
5. WinterKnight - 5 Points
6. Dominique - 4 1/2 Points
7. Fred W - 4 1/2 Points
8. John McLeod VII - 1 Point
9. Labbie - 1 Point
10. Scary Capitalist - 1/2 Point
11. TBD...

guido.man has quickly moved up the table, Sarge still not out of reach, but Bill extends his lead.....

More Questions coming soon......

Best Regards,
Luke.

---

- Luke.

<snip>
Second, what happened to the explanations portion?
</snip>

For Q39 I couldn't be doing with trying to draw my picture to explain how I derived my simultaneous equations - much easier with paper and pencil than with screen and mouse. And I got so tied up with that that I missed the obvious answer!! Oh well...

F.

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Now for the calculus point of view:

Again, we are to maximize z = -kx^2 + 80kx.

Differentiating and factoring, (dz / dx) = 2 * k *(x - 40). Setting this to zero, for the same reasons mentioned in the previous post, 2 and k cannot be zero, so 0 = x - 40, also leading to an answer of 40 hours, if indeed the critical point produces a maximum rather than a minimum or no information such as what happens when x = 0 when looking at the basic cubic function.

Either we refer to the way of thinking noted in my previous post, or, if we stick with differential calculus as Luke said would be required, we must use either the first or second derivative test.

The second derivative of z = -2k. Again, k is positive, so z'' is negative everywhere, including when x = 40, so the critical point really is a maximum.

On the other hand, using the first derivative test, we must check that the derivative changes from positive to negative in some epsilon-neighborhood of x = 40.

z' = 2 * k *(x - 40). 2k is always positive. So, we must focus on whether there is a change in x - 40. That one's clear, since anywhere to the left of 40, the line will be below the x-axis and above it to the right.

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Capitalize on this good fortune, one word can bring you round ... changes.

looks like the curve peaks at 40 hours on the work problem. The derivative is zero at 40 hours. It is a maximum

I agree with Bill's answer, but two things.
First, Luke is incorrect that differential calculus is required to answer this.
Second, what happened to the explanations portion?

If x is the number of hours of work and y is the pay, then y = k * x for some positive k because pay is proportional to hours worked. k is obviously positive based on the situation.

We are to maximize z = y * (80 - x) = k * x * (80 - x) = -kx^2 + 80kx.

When graphing z(x), we obtain a parabola that opens down. Therefore, there is no question that there is a maximum. It is found precisely at the vertex. The x-coordinate of the vertex can be found using x = -b / (2a), from the standard form z = ax^2 + bx + c.

x = -b / (2a) = -(80k) / (2 * (-k)) = 40 hours. (Noting earlier that k must be positive, based on the situation, justifies canceling the non-zero k when reducing.)

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Capitalize on this good fortune, one word can bring you round ... changes.

looks like the curve peaks at 40 hours on the work problem. The derivative is zero at 40 hours. It is a maximum

Here are two more, hopefully people find them easier....

Q39:A round table sits flush in a corner of a square room. One point on the edge of the table is 5" from one wall and 10" from the other. What is the radius of the table? (Two answers are acceptable....)

Q40:Derivative calculus required. Your job allows you to work any number of hours per week you desire. Your take home pay is proportional to the number of hours worked (no overtime). After subtracing time for sleeping and routine daily tasks you have 80 hours per week left for work and pleasure. You wish to maximize your income multiplied by the amount of pleasure time you have to enjoy it. How many hours per week should you work?

Best Regards,
Luke.

Q39: The radius of the table is 25 inches.

F.

Yes, well done Fred!
First answer to Q39: 25 inches (1 point)

There is another answer to Q39, so I will continue to advertise it as a open question....

The Math Solvers Are:
1. William Rothamel - 8 1/2 Points
2. Mr. Kevvy - 7 Points
3. WinterKnight - 5 Points
4. Sarge - 5 Points
5. Guido.Man - 5 Points
6. Dominique - 4 1/2 Points
7. Fred W - 4 1/2 Points
8. John McLeod VII - 1 Point
9. Labbie - 1 Point
10. Scary Capitalist - 1/2 Point
11. TBD...

Q39:A round table sits flush in a corner of a square room. One point on the edge of the table is 5" from one wall and 10" from the other. What is the radius of the table? (There is only one more answer for this, you may not answer with 25 inches)

Q40:Derivative calculus required. Your job allows you to work any number of hours per week you desire. Your take home pay is proportional to the number of hours worked (no overtime). After subtracing time for sleeping and routine daily tasks you have 80 hours per week left for work and pleasure. You wish to maximize your income multiplied by the amount of pleasure time you have to enjoy it. How many hours per week should you work?

Best Rgerads
Luke.

---

- Luke.

Here are two more, hopefully people find them easier....

Q39:A round table sits flush in a corner of a square room. One point on the edge of the table is 5" from one wall and 10" from the other. What is the radius of the table? (Two answers are acceptable....)

Q40:Derivative calculus required. Your job allows you to work any number of hours per week you desire. Your take home pay is proportional to the number of hours worked (no overtime). After subtracing time for sleeping and routine daily tasks you have 80 hours per week left for work and pleasure. You wish to maximize your income multiplied by the amount of pleasure time you have to enjoy it. How many hours per week should you work?

Best Regards,
Luke.

Q39: The radius of the table is 25 inches.

F.

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Here are two more, hopefully people find them easier....

Q39:A round table sits flush in a corner of a square room. One point on the edge of the table is 5" from one wall and 10" from the other. What is the radius of the table? (Two answers are acceptable....)

Q40:Derivative calculus required. Your job allows you to work any number of hours per week you desire. Your take home pay is proportional to the number of hours worked (no overtime). After subtracing time for sleeping and routine daily tasks you have 80 hours per week left for work and pleasure. You wish to maximize your income multiplied by the amount of pleasure time you have to enjoy it. How many hours per week should you work?

Best Regards,
Luke.

---

- Luke.

Congratulations! guido.man! you again have the correct answer...

And the standings are in:
1. William Rothamel - 8 1/2 Points
2. Mr. Kevvy - 7 Points
3. WinterKnight - 5 Points
4. Sarge - 5 Points
5. Guido.Man - 5 Points
6. Dominique - 4 1/2 Points
7. Fred W - 3 1/2 Points
8. John McLeod VII - 1 Point
9. Labbie - 1 Point
10. Scary Capitalist - 1/2 Point
11. TBD...

More Easier questions coming soon...

Best Regards,
Luke.

---

- Luke.

Well done to guido.man!!! for solving both questions 35 & 36!!!

Standings:
1. William Rothamel - 8 1/2 Points
2. Mr. Kevvy - 7 Points
3. WinterKnight - 5 Points
4. Sarge - 5 Points
5. Dominique - 4 1/2 Points
6. Guido.Man - 4 Points
7. Fred W - 3 1/2 Points
8. John McLeod VII - 1 Point
9. Labbie - 1 Point
10. Scary Capitalist - 1/2 Point
11. TBD...

guido.man earns himself two points for solving 35 & 36... now on four points...

And then there was one:
Q38 - A small town in Alaska is approaching winter. Because the soil will soon freeze they need to dig enough graves in the town cemetary now in anticipation of the number of deaths until the ground thaws in spring. The town's population is 1000 and it is assumed that each person has a 1% chance of dying during the winter. What is the least number of graves should they did so that the probability of having enough is at least 90%? What about 95% and 99%?

Please have a go!!!
Best Regards,
Luke.

---

- Luke.

The death of this thread is nearing, and I fear the worst, 3 questions remain.......

Q35 - Any given length of highway is equally likely to have as many potholes as any other length of equal size. The average number of potholes per mile of highway is 3. What is the probability that 2 miles of highway have 3 or fewer potholes? (Answer is not zero)

Q36 - Calculus required. There is a street of length 4. The street is initially empty. Cars then come along to fill the street until there is no space left that is large enough to park a car in. Every car is length 1. Drivers will choose a location to park at random among all possible locations left. No consideration, whether good or bad, is given to other cars. What is the expected number of cars that will be able to park?

Q38 - A small town in Alaska is approaching winter. Because the soil will soon freeze they need to dig enough graves in the town cemetary now in anticipation of the number of deaths until the ground thaws in spring. The town's population is 1000 and it is assumed that each person has a 1% chance of dying during the winter. What is the least number of graves should they did so that the probability of having enough is at least 90%? What about 95% and 99%?

As I have said before, everyone have a go!

Best Regards,
Luke.

---

- Luke.

come on guys and gals--

Lets try some more thinking on the pothole problem. In order to have 3 or fewer in two miles it has to be the one of the following joint probabilities:

first mile 0 second mile 3 or less
first mile 1 or less second mile 2 or less
first mile 2 or less second mile 1 or less
first mile 3 or less second mile 0

By summing these joint probabilities you should have the answer.

If you assumed that frequency of potholes was Gaussian distributed and that the standard deviation from the mean was 1 pothole, you could probably get the answer fairly easily.

Since the number of potholes is a discrete function (0, 1, 2, or 3) maybe a simpler reckoning other than the Normal distribution is in order.

Hmmmmmmmm...why is it the kitties and I seem to find a pothole in the first few .000001 miles? Meow........chunk........meow..........

Good to see you drop in, Mark! You seem to be referring to my problems with computers.....

Best Regards,
Luke.

---

- Luke.

Sorry, guys, but I was busy giving and then grading final exams.
After the submission of final grades (with only about 3 minutes to spare for one section), I have been recovering the last 2 days. Perhaps tomorrow I can give the problems proper thought.
BTW, Luke, yes, it is likely many people are not around, as it is a 4 day holiday weekend here in the USA.

---

Capitalize on this good fortune, one word can bring you round ... changes.

come on guys and gals--

Lets try some more thinking on the pothole problem. In order to have 3 or fewer in two miles it has to be the one of the following joint probabilities:

first mile 0 second mile 3 or less
first mile 1 or less second mile 2 or less
first mile 2 or less second mile 1 or less
first mile 3 or less second mile 0

By summing these joint probabilities you should have the answer.

If you assumed that frequency of potholes was Gaussian distributed and that the standard deviation from the mean was 1 pothole, you could probably get the answer fairly easily.

Since the number of potholes is a discrete function (0, 1, 2, or 3) maybe a simpler reckoning other than the Normal distribution is in order.

Hmmmmmmmm...why is it the kitties and I seem to find a pothole in the first few .000001 miles? Meow........chunk........meow..........

---

"Learn from yesterday. Live for today. Hope for tomorrow." Albert Einstein
"With cats." kittyman

come on guys and gals--

Lets try some more thinking on the pothole problem. In order to have 3 or fewer in two miles it has to be the one of the following joint probabilities:

first mile 0 second mile 3 or less
first mile 1 or less second mile 2 or less
first mile 2 or less second mile 1 or less
first mile 3 or less second mile 0

By summing these joint probabilities you should have the answer.

If you assumed that frequency of potholes was Gaussian distributed and that the standard deviation from the mean was 1 pothole, you could probably get the answer fairly easily.

Since the number of potholes is a discrete function (0, 1, 2, or 3) maybe a simpler reckoning other than the Normal distribution is in order.

OK, no one is answering these three, are they too hard? or is no one around? last try to keep this thread alive:

Q35 - Any given length of highway is equally likely to have as many potholes as any other length of equal size. The average number of potholes per mile of highway is 3. What is the probability that 2 miles of highway have 3 or fewer potholes?

Q36 - Calculus required. There is a street of length 4. The street is initially empty. Cars then come along to fill the street until there is no space left that is large enough to park a car in. Every car is length 1. Drivers will choose a location to park at random among all possible locations left. No consideration, whether good or bad, is given to other cars. What is the expected number of cars that will be able to park?

Q38 - A small town in Alaska is approaching winter. Because the soil will soon freeze they need to dig enough graves in the town cemetary now in anticipation of the number of deaths until the ground thaws in spring. The town's population is 1000 and it is assumed that each person has a 1% chance of dying during the winter. What is the least number of graves should they did so that the probability of having enough is at least 90%? What about 95% and 99%?

Best Regards,
Luke.

---

- Luke.

Luke,
Why haven't you accepted my answer to Q35 of none. Don't you realise you are talking to an expert here.
I live within four miles of Blackburn, Lancashire.
A day in the life
40 years later, Latest report.

WinterKnight, zero is not the correct answer....... so I cannot give you the point.....

Best Regards,
Luke.

EDIT: here are the current questions....
Q35 - Any given length of highway is equally likely to have as many potholes as any other length of equal size. The average number of potholes per mile of highway is 3. What is the probability that 2 miles of highway have 3 or fewer potholes?

Q36 - Calculus required. There is a street of length 4. The street is initially empty. Cars then come along to fill the street until there is no space left that is large enough to park a car in. Every car is length 1. Drivers will choose a location to park at random among all possible locations left. No consideration, whether good or bad, is given to other cars. What is the expected number of cars that will be able to park?

Q38 - A small town in Alaska is approaching winter. Because the soil will soon freeze they need to dig enough graves in the town cemetary now in anticipation of the number of deaths until the ground thaws in spring. The town's population is 1000 and it is assumed that each person has a 1% chance of dying during the winter. What is the least number of graves should they did so that the probability of having enough is at least 90%? What about 95% and 99%?

---

- Luke.

Luke,
Why haven't you accepted my answer to Q35 of none. Don't you realise you are talking to an expert here.
I live within four miles of Blackburn, Lancashire.
A day in the life
40 years later, Latest report.

Q35 - Any given length of highway is equally likely to have as many potholes as any other length of equal size. The average number of potholes per mile of highway is 3. What is the probability that 2 miles of highway have 3 or fewer potholes?

Seriously - 100% probability, that somewhere in California or Florida, there are 2 miles of road with 3 or fewer potholes. LOL

"For a given length of highway of 2 miles", with an average of 3 potholes per mile, the probability of 3 or less potholes, is ZERO.

Wrong, the answer is close to zero, but it is not zero itself....
Try Again!!!

Best Regards,
Luke.

Thinking out loud to try to stimulate others.

Hint: I think that you have to assume that the Probability Curve here for the potholes is normally distributed. ( Gaussian--Bell Shaped Curve) and then figure out how far you would have to be on the tail of the curve to give the required number of potholes or less. Then take that area under the distribution curve to see the probability. You could probably use a table for the Normal distribution or maybe find a tool on the Internet. The only thing that bothers me about this line of thinking is that we don't have the standard deviation from the mean--this may suggest that simpler methods are called for.

Conversely, It may be more complicated since you really have a joint probability since there are two stretches involved, and the average is for one stretch.

I am sure this is no help but see if someone can solve Lukes problem anyway .

Cheers rite mates,

Bill