Optimization using derivatives (word problems)

[cobainage1994]

I have a math project, me and my partner are completely stumped.

1. you are going to fly a plane 900 miles at a constant mph. you know that if you fly 120 mph you will get 10 miles to the gallon and that for every 1 mph over 120 mph your mpg is decreased by .04 mpg. the cost of fuel is $4.75 per gallon. the cost to keep the airplane in the air is $52 per hour. the plane can fly a minimum of 90mph and a maximum of 180mph. this is a closed interval test [90,180]
To the closest hundredth, find the mph to fly to minimize the cost. also, what is the cost? do not do any rounding off till the final answer.

2. you are a lifeguard and are standing on the shore of a calm lake. you see a person in need of help who is 500 meters directly down the shore and 250 meters out. you can run at a rate of 9 meters per second and can swim at the rate of 3 meters per second. how far down the shore should you run before entering the water to swim in orderto minimize the time to get to the person in need of help? how long will it take you to gett there? (tenths)
this is an open interval, use 10 and 100 as test numbers with the first derivative

For the first problem we think the equation is C= 4.75(900/10-.04x)+52(900/120+x). We are not sure though

For the second problem.. We got nothing.. Thanks in advance for any help!

 

[mmm4444bot]

… for every 1 mph over 120 mph your mpg is decreased by .04 mpg …

I note that the author of this exercise made no statements about what happens to the mileage when the speed is less than 120 mph.

Therefore, I would proceed only after assuming that the answer to "how fast" will be greater than 120. If this assumption works out, then we can ignore what happens when x is from 90 through 120.

'

… we think the equation is C= 4.75(900/10-.04x)+52(900/120+x) …

I think it's important to state up-front the meaning of variables we choose, to help others understand what we're doing.

Let x = the speed of the plane in miles per hour (mph)

Let C(x) = the operating cost to fly the plane 900 miles at speed x

Your algebraic definition for C(x) is very close.

See if you can find your two mistakes, after thinking about the following.

What is the operating cost to fly this plane 900 miles at 150 mph?

x = 150

Distance/Rate = Time

900/150 = 6

This flight takes 6 hours.

6(52) = 312

The non-fuel operating cost is $312.00 .

180 - 150 = 30

The plane flys 30 mph above the 120 mph threshold.

0.04(30) = 1.2

10 - 1.2 = 8.8

The mileage is 8.8 miles per gallon.

900/8.8 = 102.272727…

102.27 gallons of fuel are burned.

4.75(102.27) = 485.795454…

The fuel cost is $485.80 .

312 + 485.8 = 797.80

The operating cost to fly the plane 900 miles at 150 mph is $797.80 .

In other words, C(150) = 797.80 .

Your definition gives C(150) = 1242.08 .

HINT: If you cannot find your errors in logic, then focus on the following two expressions within your definition for C(x).

(900/10-.04x)

(900/120+x)

By the way, you have not properly typed these expressions because -- without grouping symbols -- they are ambiguous.

Here is how you should type them.

[900/(10 - 0.04x)]

[900/(120 + x)]

Let us know if you still cannot find the flaws in these two expressions.

After you've got the correct expression for the definition of function C, we can move on to the derivative phase of this exercise.

 

[mmm4444bot]

… For the second problem.. We got nothing …

To get started on the second exercise, here's a picture to look at. (Did you draw one?)

Double-click the image to expand it.

[attachment=0:2r7b6crp]Lifeguard.JPG[/attachment:2r7b6crp]

The green line segment represents some distance covered by running, and the red line segment represents some distance covered by swimming.

The exercise asks for the length of the green segment such that the entire distance is covered in the least amount of time.

So, let x represent the length of the green segment.

Use the Pythagorean Theorem to express the length of the red segment in terms of x.

To write the function that gives the total time in terms of x, remember that elapsed time can be expressed as the ratio of distance-covered over the rate.

T = D/R

Please show your work, if you need more help.

 

[BigGlenntheHeavy]

Re: Optimization using derivatives (world problems)

2nd Problem. Refer to the above diagram of mmm4444bot.

Let S(swim) = hypotenuse of the above right triangle (red line) and R(run) = other leg of right triangle (base line).

Then 500-R = green line. Given: dS/dt = 3 and dR/dt = 9.

Observing the right triangle, we have S^2 = (250)^2 + R^2 which implies that 2S(dS/dt) = 2R(dR/dt).

Thus 3S = 9R or S = 3R. Now (3R)^2 =(250)^2+R^2 implies R = 88.3883m, the S = 3R = 265.165m and 500-R = 411.612m.

Ergo the man runs 411.612m and then swims 265.165m to rescue the damsel in distress.

Time: Since he can run 9m per sec and swim 3m per sec, it will take him 411.612/9+265.165/3 or 134.123 sec.or about 2.24 min to come to the aid of Polly Purebred.

 

[fasteddie65]

Re: Optimization using derivatives (world problems)

The second problem about find the point on the shore is one of my favorite types of optimization problems. There is usually one like that in every calculus text. The variations involve oil pipelines under the desert, as well as island problems like this one.

 

[cobainage1994]

Re: Optimization using derivatives (world problems)

I talked to my calc prof, and he said that

C(x)= 4.75[900/(10 - 0.04x)] + 52[900/(120 + x)] is correct. X being every mile over 120. Now I just need to set the C`(x) to 0 and solve for X I believe.

Thank you for everyone's help so far!!

 

[cobainage1994]

Re: Optimization using derivatives (world problems)

Alright, I finished the first problem, but am still having great troubles with the second one. I understand the picture and whats going on, but I do not know how to come up with the proper equation.

 

[mmm4444bot]

… X being every mile over 120 …

Good grief! If the plane flys 900 miles, then it flies 780 miles over 120 miles, regardless of its speed.

I sure hope that you're thinking "X being the number of mph over 120 mph", instead. (You typed something else entirely.) Can you understand the difference?

From your original post, I had to GUESS what X represents because you never said. So, I chose to let X represent the speed of the plane because that is exactly what the exercise is asking for.

'

… Now I just need to set the C`(x) to 0 and solve for X I believe.

You need to do this, but this is not "just" what you need to do.

If you use my definition of X, then the zero of C'(X) is the answer.

The exercise asks for the speed of the plane; it does not ask for the number of miles per hour above 120 mph. After you find the zero of C'(x), you will need to do an additional calculation to get the answer to the exercise.

Please let me know if you don't understand what I'm talking about.

If you would like to compare answers, then post your result, and I will complete this exercise also.

 

[cobainage1994]

Re: Optimization using derivatives (world problems)

I sure hope that you're thinking "X being the number of mph over 120 mph", instead. (You typed something else entirely.) Can you understand the difference?

Yes, I meant to say x being the number of mph over 120mph. Sorry for being so unclear. These problems have just been having me flustered, they are light years ahead as far as difficulty as to what we generally do in the class.


Alright, I have the Correct equation and derivative, according to my prof.

I changed a few things up from what I had originally stated. I am letting X equal the actual MPh as opposed to mph over 120.

So C(x)= (900/x)(52)+(900/14.8-.04X)(4.75)= (46800x^-1)+ (4275/(14.8-.04x))

I have C`(x)= (46800/x^2)+(171/(14.8-.04x))^2


Like I said, it has been confirmed to me that both of these are correct.

So I need to solve for X, and to do this I did..

0=(46800/x^2)+(171/(14.8-.04x))^2

(46800/x^2)=(171/(14.8-.04x))^2

46800(14.8-.04x)^2=171x^2

(14.8-.04x)^2= .0036538462x^2

219.09-1.184x+.0016x^2=.0036538462x^2

.0020538462x^2+1.184x-219.09

X is approx 147.34

Now according to this, the speed to minimize costs should be 147.34MPh However, when I test 90,147.34, and 180. C(180)<C(147.34)

So I must have done something wrong, I can not find the error in my equation though.





I am pretty sure I have the second problem correct.

He should run 411.612 meters before entering water, and it will take him 2 mins 14 seconds to reach the victim.

 

[mmm4444bot]

… These problems have just been having me flustered, they are light years ahead as far as difficulty as to what we generally do …

If you determined both of your versions of C(x), as well as C`(x), on your own, then you are light years ahead of most calculus students who seek tutoring where I've worked. Furthermore, simply comprehending the problem description and how the derivative applies to the solution in this exercise places you above most human animals on the planet.

'

C(x)= (900/x)(52)+(900/14.8-.04X)(4.75)= (46800x^-1)+ (4275/(14.8-.04x))

C`(x)= (-46800/x^2)+(171/(14.8-.04x))^2 (Typographical error)

Your results for C(x) and C`(x) match mine.

Note: When typing, the grouping symbols are only needed around the binomial denominator.

C(x) = 46800/x + 4275/(14.8 - 0.04x)

C`(x)= -46800/x^2 + 171/(14.8 - 0.04x)^2

'

… 0=(-46800/x^2)+(171/(14.8-.04x))^2 … (Typographical error)

… 219.04-1.184x+.0016x^2=.0036538462x^2 (Insignificant rounding error corrected here and below)

.0020538462x^2+1.184x-219.04

X is approx 147.34 …

… However, when I test … C(180)<C(147.34) …

… I can not find the error in my equation …

147.34 miles per hour is the correct answer. The equations that you posted are all correct (as edited by me). Check your arithmetic in the function evaluations:

C(147.34) = 797.62

C(180) = 822.50

'

… I am pretty sure I have the second problem correct …

… He should run 411.612 meters …

I am pretty sure, too, since your results match the answers posted earlier in the Big Heavy Glenn solution.

(I would report 411.6 meters; that extra 12/1000ths of a meter is only 12 mm, and I do not believe -- off the top of my head -- that it's enough to add another second to the travel time.)

 

[cobainage1994]

Re:

… These problems have just been having me flustered, they are light years ahead as far as difficulty as to what we generally do …

If you determined both of your versions of C(x), as well as C`(x), on your own, then you are light years ahead of most calculus students who seek tutoring where I've worked. Furthermore, simply comprehending the problem description and how the derivative applies to the solution in this exercise places you above most human animals on the planet.

'

I would be a liar if I said I found those equations 100% by myself. I am just lucky enough to know a few math majors, and this problem even stumped a few of them :?

I am in is Pre-calculus. Could just be me, but this problem felt like flat out calculus

Note: When typing, the grouping symbols are only needed around the binomial denominator.

C(x) = 46800/x + 4275/(14.8 - 0.04x)

C`(x)= -46800/x^2 + 171/(14.8 - 0.04x)^2

'
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I think the amount of parentheses I had in my C(x) is what screw me up actually. When I typed the test numbers into the equations you supplied above got the correct answer. So thank you very much for that!

… 0=(-46800/x^2)+(171/(14.8-.04x))^2 … (Typographical error)

… 219.04-1.184x+.0016x^2=.0036538462x^2 (Insignificant rounding error corrected here and below)

.0020538462x^2+1.184x-219.04

X is approx 147.34 …

Yeah, both of those errors were typos, had the - and .04 on my worksheet.

147.34 miles per hour is the correct answer. The equations that you posted are all correct (as edited by me). Check your arithmetic in the function evaluations:

C(147.34) = 797.62

C(180) = 822.50

'

All in all, thank you SO much for all your help!!

 

[mmm4444bot]

… I am in is Pre-calculus …

As Spaceman Spiff (who generally materializes in Calvin's math class) would say, "Gadzooks!"

That be one advanced pre-calculus course! :shock:

 

[cobainage1994]

Re:

… I am in is Pre-calculus …

As Spaceman Spiff (who generally materializes in Calvin's math class) would say, "Gadzooks!"

That be one advanced pre-calculus course! :shock:

You would think so based on that problem. But like I said, that problem was ridiculous compared to the other optimization stuff we had.

Here is the example problem he gave us before he handed us the worksheet that contained the above problems.


(note: this is the condensed version)
Theatre owner sells 250 tickets if she charges $5 a ticket. She loses 10 customers for every dollar the price increases. Find what she should charge to maximize revenue

There is no way THAT^ problem can prep you for the problems in the original post.

 

[BigGlenntheHeavy]

f(x) = (5+x)(250-10x), find its derivative and set to zero and you'll have the max.

 

[mmm4444bot]

… You would think so based on that problem …

If your class were using technology (eg: computer software, graphing calculator) to determine the coordinates at the minimum value on the graph of function C, then this exercise does not seem unusual (to me) for a precalculus course.

My surprise comes from learning that your class is determining derivatives of rational functions. I think that is fairly advanced for precalculus.

(Perhaps, you'll be working with differential equations, if you take introductory calculus at the same school. :wink: )