Difficult Functions Question from 1996 AHSME #25

[347670531]

-36 in the third line should be -9

[lamestguyintown]

cant he pronounce three?

[SeanieG69]

yes lagrange multipliers would work, they can solve most constrained optimisation problems. however for such a simple constraint we can parametrise our constraint curve in terms of a single variable (theta) and apply methods of single-variable calculus, which are generally simpler.

[colplaya7]

wouldn't lagrange multipliers work?

[hacksign23]

dat is duh gut one!

[maths486]

nice solution heissenburg. i've also tried using pure algebra using the discriminant of the quadratic equation

[DoubleDutchBust]

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yekalieleeyno DDifficult Functions QuestionAHSME #25

[DoubleDutchBust]

1971 Mel Blanc American Express TV commercialDifficult Functions Question from 1996 AHSME #25

donyleeTwentyFive#18

cabcdefghijklmnop 16+1:08 4 c

cinco de mayo

donylee sean

[iridethewave]

(continued from below)

... when the two values are approximately equivalent so at theta=pi/4

solving out gives(simplified) = (12 + 16)sqrt(2) + 33 = 29sqrt(2) + 33 =

72.6, which is close enough an approximation given our inference

[Heissenburger]

good idea, but use the R-formula (R=sqrt(24^2 + 32^2) = 40) to get

it into an expression with one theta varying. so 24cos(theta) + 32sin(theta) = 40 sin(theta + alpha)

[alpha = invtan 24/32, but not important] then obviously the max

of this is 40, and overall max is 40 + 33 = 73.

[iridethewave]

i agree heissenburger, that's quite an effective solution. I now feel like a numerical analysis heau.

[iridethewave]

I solved this with a much more efficient method:

represent x and y in terms of cos and sin, you understand from the equation of the circle that each x can be represented as the radius' x distance from 0 + cos of some variable theta, doing the same with y gives

1) 3x + 4y = 24cos(theta) + 32sin(theta) + 33

substituting cos(theta) = sin(theta + pi/2) into (1) we get

2) 24sin(theta + pi/2) + 32sin(theta) + 33 = 3x +4y

through induction we can infer that said expression is maximized

[bobthebuilder123451]

@iridethewave Oh, dang, now I know why parametric functions are useful=)

[tenkuhito]

You spent too much energy to speak for such an easy problem.

[Jdelozier828]

Why couldn't you simply substitute the value of y in 3x + 4y with the equation of the circle solved for y, then take the derivative with respect to x and solve for zeroes to find minimums and maximums?

[mbabitt]

When I got that I got something way off: 365

:huh: This isn't a very neat method, but it's neater than derivative-bashing.

[iridethewave]

Since these are AMC problems they always have solutions which can be obtained without the use of calculus and this is what donny is doing. I guess he wants to solve it within the bounds

[senox88]

wow.

u serious.

i like it!

good luck

dony.

hope one day MIT, you're in.

[scott01019]

Cool I never understood how to do those types of problems until watching this video. Awesome job explaining this. Thank you :).

[XUnrealpkI]

this didn't work out so, i went in reverse from your answers and accidentally figured out the max value of t for x1 and y1 occur when d/dt(line integral) is maximum. Why is the maximum values of x1 and y1 located when d/dt(line integral) is maximum, it doesn't make sense.

[XUnrealpkI]

hey donylee, i tried to attempt this question with line integrals and the fund theory of calc(2).

The way i understood the question was find the maximum value of z= 3x + 4y under the circle

(x-7)^2 + (y-3)^2 = 64, since i couldn't parametrize this circle for the line integral, i shifted the plane, so z = 3x1+4y1-33 under the curve x1^2+y1^2=64. I think you get what i'm trying to do. I thought the values of x1 and y1 would occur when d/dt(line integral from 0 to t) = 0.

[Wuzzi55]

lmfao! what is this? grade 10 math??

Dude, please make a video about hardcore math questions, that someone would might face in university. Then it's worth watching more u know...

But still good job :D

[bkjoelover]

Are you serious?

[Entertainmentwf]

lol, what are you talking about, that's a NCS/MAA level question lol, it isn't grade 10 math. If you encounter a problem like this outside of high school you'd be expected to prove everything. It is a tricky question, and he is simply solving this problem cause there are many ways to go wrong on this one. No grade 10 mathematics course teaches those axioms he used (a couple I saw there). There are many ways to solve this problem but, he used a tricky way which is a good way to attack this.

[ihateuutube]

well, this is from a competition that 10th graders take.... so kinda.

[solo2wolf]

i would have differentiated the circle equation and equated that to the gradient we know we want and thus you have two equations, that and the original and solve by substitution. your method was beautiful though.great stuff as always.

[donylee]

Yup, I see what you mean. Though I must ask how did you get the gradient we know we want.

Probably there's a theorem, but it seems that to maximize '3x + 4y' we need to turn this into a y = mx + c form, compare it with the circle and then equate the derivative of the circle to m.

[ataggake]

As a Maths teacher myself (with students at this level) I found this very interesting. Something to think about - does the introduction of "c" and especially "a" actually help here, or does it confuse?

Once the concept of tangent is understood (and thus we know the gradient of the line from the centre to the solution point), I would tend to go straight to (y-3) = 4/3 (x-7) substituted into the equation of the circle.

Also interesting to compare with the calc. solution.

[ataggake]

(Very sorry for duplicate!!)

[donylee]

Hello ataggake,

I was reading your suggestion. The reason for introducing the 'c' and 'a' is to find the function y=y(x) that can be intersected with the circle to get our desired point, the point which maximized our original function 3x+4y.

As for your method, I got to question how did you get the equation (y-3) = 4/3 (x-7) to begin with. You are ALREADY saying that the line from the centre with gradient 4/3 will give the maximum value. I can't see this, must be some theorem.

[ataggake]

You know, on reflection, I thought it really is helpful to include c. It really depends on your students. Showing them the y-intercept is a good concrete illustration of what's happening.

As for your question about where the equation comes from, I'm just saying that since you're looking at a tangent, the line from the centre has gradient -1/(gradient of tangent), because perpendicular. No big deal.

I actually thought it was a very clear bit of teaching, well done :)

[ataggake]

You know, on reflection, I thought it really is helpful to include c. It really depends on your students. Showing them the y-intercept is a good concrete illustration of what's happening.

As for your question about where the equation comes from, I'm just saying that since you're looking at a tangent, the line from the centre has gradient -1/(gradient of tangent), because perpendicular. No big deal.

I actually thought it was a very clear bit of teaching, well done :)

[jonahansen]

Way to go. There is a typo when you complete the square for the y; you subtract 36 instead of 9.

[gamesguru]

I think Lagrange multipliers are more well suited for this problem. This geometric representation becomes a bit less inconvenient once you learn about Lagrange multipliers. But this is a good way to demonstrate a point to first year calculus students. Good Job!

[ihateuutube]

this is ashme. students are not expected to use calculus and everything can be done precalculus. this prob seems pretty easy for #25, straightforward algebra.

[YasinIsse]

Why do you make this videos?

[donylee]

To teach mathematics to those with the desire, patience and curiosity to learn.

[timwu87]

A formula that helps:

The distance from a point(x_1,y_1) to a line ax+by+c=0 is |ax_1+by_1+c|/sqrt(a^2+b^2).

In this case, we have line 3x+4y-k=0, and point (7,3)(which is the center of the circle). So to solve for tangency,

|3*7+4*3-k|/sqrt(3^2+4^2)=8, |33-k|=40, k=-7 or 73, and 73 is our answer.

[bertazoid]

Good comment, thanks

[12354blahbla]

Another way to solve it using trigonometry:

using some trigonometric functions you can describe the situation as:

(4(3+8sin(x))+(3(7+8cos(x)) and graph the equation

basically: 4(y)+3(x)

where (x,y) is a point on the circle

this works because sin=y/r and cos=x/r so sin+cos=y+x

[donylee]

Thanks everyone for your enthusiasm and interest in the question. Yes, this can be easily done using Lagrange multipliers or the Hessian Matrix, something which I don't know, like josembi12 suggested.

Though I must say, AMC are meant for high school non-calculus students. Can't imagine them partial differentiating the function.

Still, I appreciate the discussion. :-)

[josembi]

oops, may be I lopped off..went too far. Didn't realise it. My bad. I should have directed it to Calculus students.

[bertazoid]

agreed...I think its obvious what kind of level this question is aimed at, so Lagrange multipliers are out of bounds.

It might not be that advanced in general, but for the level of students this question is aimed at it is a difficult one...I remember struggling with similar questions before. Anyway the video was really well explained and thourough...nice job..admirable to make the videos for these students...I wish they were here when I was there! :) Cheers mate!

[Claudiamaths]

I dont see whats wrong with Lagrange multipliers. Obviously been buoyant in disposition and open to learning is what makes you stand out of the crowd.

If its indicated that you cant cross a chasm with two small jumps, dont fear to make a big jump. Dont be intimidated by the knowledge of others. Learn from them.

[TheShark909]

Very nice, it makes sense quite well. You nearly lost me in some points, but do to your summarization I was able to grasp at what you were getting at. Nice job!

[josembi12]

- Its also important to underline the fact that if we had the fn without the constraint, we would have used the Hessian Matrix to compute for Relative MAX/MIN and Absolute MAX/MIN ⇔ ∃.

H(f) = ∂^2f/∂x^2 ........Hv to rush and do some stuff will finish this later.

by Josembi Italy

[Claudiamaths]

Hey josembi12, you promised to show us the Hessian method...what happened? We are still waiting.

Claudia

[josembi12]

Using Lagrange multipliers,

L(x,y,φ) = f(x,y)+ φg(x,y)

L(x,y,φ) = 3x+4y + φ(x^2-14x+y^2-6=0)

Get the ∇.Partial deriv of x,y,φ.(f∈ℝ^2)

∇L(x,y,φ)= ∂L/∂x = 3+2xφ-14φ ∂L/∂y = 4+2yφ  ∂L/∂φ = x^2-14x+y^2-6

Computing φ (lagrange multiplier), we'll be able to get values of x,y, and even z if f ∈ ℝ^3.

Subs't these values in the fn, we get Absolute MAX and Absolute MIN.

[JWHooper3581]

at 1:16, the third equation should be (x-7)^2-49+(y-3)^2-9=6, so replace 9 over 36.

[JWHooper3581]

this can be solved using Cauchy-Schwartz inequality, like senlin1990 said.

[jaysin1947]

this is suoerb..this helped me a lot..thanks mr.donyleen n i got the answer as 73,is that correct?

[kooltrainer]

noice.. cant you solve this using derivative? i just wana know how .. =)

[ToOtLoL]

Nice this really helps me !

[eznamee]

hey donny. i was wondering if u could do antiderivatives by sustitutions. am in calculus 1 right now and basically stuck. thx.

[donylee]

Hello Eznamee,

I would really like to help you. Unfortunately, I'm currently doing a series of videos on Fourier Analysis followed by Surface Integrals. Surface Integrals will most likely show up on your Calculus 2 course.

If I have the time, I'll try my best to do some videos on integrating by substitutions.

('Antiderivatives' is actually the proper term but nobody uses that now. Hehe.)

[senlin1990]

use substitution: a=x-7 and b=y-3, then you have a^2+b^2 = 64 and you want to find the max of 3a+4b+33, which can easily be found using:

(9+16)(a^2+b^2)>=(3a+4b)^2

so 3a+4b>=40 and hence x+y=3a+4b+33>=73

[donylee]

A written solution is found on my webpage. Check the description for the link.

I would like to also thank everyone for their enthusiasm in this problem.

[aioqwe]

I got 75.4

[donylee]

Hey aioqwe,

Thanks for participating in the discussion. I have triple checked my answer, as well as checking the answer from 'First Steps For Math Olympians'. It is 73, using this method.

[aioqwe]

Oh yeah.. I went back and realized that replacing my answer into 3x+4y, I placed my y value into 3x and my y value into 4y.

[mediteight]

Enthusiastic explanation :) - neat video, thanks. I like your approach for high school maths a lot but adb4 does have a good point. The Lagrange multiplier method can be applied here with far less thought - (though this might not be a good thing for your purposes!)

In case you're interested I'll try out adb4's approach ...

[adb4]

use the method of lagrange multipliers. Much less thought for the same result.

[donylee]

Woah, looks like that's some advance stuff. Honestly speaking, I do not know lagrange multipliers and Cauchy's inequality well enough for it to be applied here.

Guess it shows AMC questions are open to many approaches.

[adb4]

well the thing is, the method of lagrange multipliers also allows you to use more than one constraint. you can use as many constraints as you have variables, if I am not mistaken. It also allows this sort of "maximization / minimization with constraints" problem to be solved on a larger class of problems, where essentially if the constraints are differentiable, (even numerically) then a solution can be found.

[zsyed12345678]

Answer is 73