What is the trick to do this problem w/o calculus

[Steven G]

Possibly this problem should be in thew beginning algebra sub forum as I was told that a middle school student should be able to solve this. (Where is Khan when you need him?!)

Minimize \(\displaystyle \sqrt{x^2+4} + \sqrt{y^2 + 9}~ subject~ to~ x+y=12\)

No Calculus allowed.

 

[The Highlander]

Possibly this problem should be in thew beginning algebra sub forum as I was told that a middle school student should be able to solve this. (Where is Khan when you need him?!)

Minimize \(\displaystyle \sqrt{x^2+4} + \sqrt{y^2 + 9}~ subject~ to~ x+y=12\)

No Calculus allowed.

\(\displaystyle x=5, y=7\).

 

[topsquark]

\(\displaystyle x=5, y=7\).

Are you sure about that? (The corner is looking very inviting right now...)

The solution isn't too bad, but I could only think to graph it. If the number hadn't come out nice it would have been too hard.

And I wouldn't say that this is middle school level at all!

-Dan

 

[canvas]

drawing a rectangle with a width: 12 = x + y and height: 2 & 3, using the pythagorean theorem we get that hypotenuses are sqrt(x^2+4) and sqrt(y^2+9)

 

[Dr.Peterson]

drawing a rectangle with a width: 12 = x + y and height: 2 & 3, using the pythagorean theorem we get that hypotenuses are sqrt(x^2+4) and sqrt(y^2+9)

Very nice hint -- it let me solve the problem quickly (and see that the answer is not 5 and 7, but close), but left most of the thinking to me.

 

[canvas]

Very nice hint -- it let me solve the problem quickly (and see that the answer is not 5 and 7, but close), but left most of the thinking to me.

I think the question is about to minimize the expression, we don't need to get exact values o

 

[canvas]

of x and y, but the value of the expression
[math]\sqrt{x^2+4}+\sqrt{y^2+9}\geq\sqrt{x^2+y^2+4+9}\geq13\\\\[/math]

 

[canvas]

missed:
[math]...\geq\sqrt{(x+y)^2+(2+3)^2}=\sqrt{12^2+5^2}=13[/math]

 

[Dr.Peterson]

I think the question is about to minimize the expression, we don't need to get exact values of x and y

Are you saying that you yourself don't see the quick geometrical solution?

Exact values are easy to get, so surely that is what is intended.

Of course, I don't think that most middle school students would even understand what the question means; but all you need to know, beyond the meaning of "minimize" and "subject to", are Pythagoras and an idea about the shortest distance between two points.

 

[The Highlander]

Are you sure about that? (The corner is looking very inviting right now...)

No, I’m not the least bit certain about it! lol.

I just had a quick go at solving it using Trial & Error/Iteration along with a soupçon of IT added into the mix. (See attached spreadsheet.xlsx) Maybe I should have looked beyond 10 decimal places?

If I’m due a session in the corner then I’ll happily sit there for a while but my answer was “close”, was it not?
Don’t I get off for at least trying and a near miss?

I’m afraid I just don’t get @canvas' 'reasoning' at all. In fact, it’s confusing me to the point where I’m going round in ever-decreasing circles! @Dr.Peterson calling it a “Very nice hint” makes me think it must be “right” & I’m just not seeing something obvious but I can’t get it to hold water:-

drawing a rectangle with a width: 12 = x + y and height: 2 & 3, using the pythagorean theorem we get that hypotenuses are sqrt(x^2+4) and sqrt(y^2+9)

I’m confused, firstly, by the use of “&” followed by “and” (as highlighted above)! Do these mean “+” (as in adding them together) or do they indicate separate ‘rectangles’? Either way I just don’t see how you can end up with “hypotenuse(s?)” that are: “sqrt(x^2+4) and sqrt(y^2+9)”.

Using any interpretation of canvas' postulation I always end up with ‘extra terms’ (eg: \(\displaystyle 2xy\)) that prevent arrival at the result claimed. (Edit: Just noticed, after posting, there is a square root sign missing in the the third figure where "h=..."; mea (maxima) culpa.?)

It looks to me as if s/he may be equating things that aren’t true (though I’m not sure where).

Does their analysis depend on: \(\displaystyle \sqrt{x^2+4}+\sqrt{y^2+9}=\sqrt{x^2+4+y^2+9}\) being ‘used’ somewhere?
That, of course, is patently untrue as: \(\displaystyle \sqrt{a+b}\neq\sqrt{a}+\sqrt{b}\), eg: \(\displaystyle \sqrt{4+9}\neq\sqrt{4}+\sqrt{9}\text{, ie: }\sqrt{13}\neq 2+3\)!

Are you saying that you yourself don't see the quick geometrical solution?

all you need to know, beyond the meaning of "minimize" and "subject to", are Pythagoras and an idea about the shortest distance between two points.

No, I, for one, certainly don’t "see" it! I’m now ‘stuck’ on trying to get canvas’ diagonals to work out, lol

Would you please be good enough to present your solution, Dr.P.?

I (among others, I'm sure) would very much like to see it (so I can get some sleep! )

 

[Steven G]

@The Highlander, I too am confused as what Dr Peterson is talking about.
I would make one comment on your work. You wrote (x+y)^2 in a number of places and foiled it out. It might have been better to say that (x+y)^2 = 12^2 = 144

Also, canvas wrote sqrt(x^2+4) + sqrt(y^2+9) > sqrt(x^2+4 + y^2 + 9), not sqrt(x^2+4) + sqrt(y^2+9) = sqrt(x^2+4 + y^2 + 9)

 

[Dr.Peterson]

Would you please be good enough to present your solution, @Dr.P.?

I too am confused as what Dr Peterson is talking about.

Draw a 12 by 5 rectangle, and in the lower left, make a rectangle x by 2; its diagonal will be [imath]\sqrt{x^2+4}[/imath]. Make a y by 3 rectangle from the upper right of the small rectangle you made, to the upper right of the big rectangle. Its diagonal will be [imath]\sqrt{y^2+9}[/imath].

To minimize the sum of those two diagonals, you are looking for the shortest path from the lower left to the upper right,, which will be the diagonal of the large rectangle. What must x and y be to make that happen?

 

[The Highlander]

@The Highlander, I too am confused as what Dr Peterson is talking about.

Perhaps he will be good enough to come back and explain in more detail; that’s my hope anyway.
(Edit: I see he has now done so; thank you for that @Dr.Peterson.)

I would make one comment on your work. You wrote (x+y)^2 in a number of places and foiled it out. It might have been better to say that (x+y)^2 = 12^2 = 144

That, of course would be a perfectly valid option, but, surely, if (x+y)² is replaced throughout by 144 (ie: 12²) then all traces of the (x²+4) & (y²+9) just disappear and the three triangles I drew just become 5,12,13; 2,12,\(\displaystyle \sqrt{148}\); & 3,12,\(\displaystyle \sqrt{153}\) triangles respectively which rather defeats the 'purpose' of trying to express the hypotenuse(s) in terms of (x²+4) & (y²+9) (or the square roots of those expressions)?

Also, canvas wrote sqrt(x^2+4) + sqrt(y^2+9) > sqrt(x^2+4 + y^2 + 9), not sqrt(x^2+4) + sqrt(y^2+9) = sqrt(x^2+4 + y^2 + 9)

Indeed, there is no disputing that, however, that is, partly, why I said that I was “not sure where” canvas might have been equating \(\displaystyle \sqrt{x^2+4}+\sqrt{y^2+9}\textbf{ to }\sqrt{x^2+4+y^2+9}\) but I couldn't escape the suspicion that such an equivalence had been assumed somewhere in their reasoning.
I am happy to accept that I may be completely wrong in that respect but it's just not clear (to me, at least) from what was posted how the initial conclusion (that any "hypotenuses are \(\displaystyle \sqrt{(x^2+4)}\) "and" \(\displaystyle \sqrt{(y^2+9)}\)") is reached based on the subsequent algebraic manipulations presented.
(Especially the line: "\(\displaystyle \sqrt{x^2+4}+\sqrt{y^2+9}\geq\sqrt{x^2+y^2+4+9}\geq13\)"; where does that come from?)

 

[Steven G]

(Especially the line: "\(\displaystyle \sqrt{x^2+4}+\sqrt{y^2+9}\geq\sqrt{x^2+y^2+4+9}\geq13\)"; where does that come from?)

It does give a lower bound for \(\displaystyle \sqrt{x^2+4}+\sqrt{y^2+9}\)

 

[Steven G]

Draw a 12 by 5 rectangle, and in the lower left, make a rectangle x by 2; its diagonal will be [imath]\sqrt{x^2+4}[/imath]. Make a y by 3 rectangle from the upper right of the small rectangle you made, to the upper right of the big rectangle. Its diagonal will be [imath]\sqrt{y^2+9}[/imath].

To minimize the sum of those two diagonals, you are looking for the shortest path from the lower left to the upper right,, which will be the diagonal of the large rectangle. What must x and y be to make that happen?

Thanks for your explanation. I understand how to solve this problem with complete understanding because of your post.
Have a great Thanksgiving tomorrow!