Hardest maths questions - minimisation

[@MarinHristov-n8m]

We can't just asume that x

[@tjerkharkema7378]

clever lassy, I love your way of thinking outside of the box, thnx for the guidance, TJ

[@lightsyagami8958]

Hello, really, I must thank you because the way to solve this exercise helps me in the mathematical Olympics, I did my best and I hope to win a gold medal, I don't know if I could solve the exercise without this method, thank you very much.

[@TheSabian321]

I solved a similar problem years ago. The difference is that the sum of the hypotenuses and bases is a constant and you have to maximize the sum of heights. The answer to that problem would also solve this one essentially.

[@MGSchmahl]

I tried this too, substituting p=x, q=y-x, r=z-y, and s=10-z. Using Lagrange multipliers, I got q=±2p, r=±p, s=±3p, and p=10/7, ±10, ±2, or ±10/3. The original expression reduces to 7√(p²+1). The minimum of √149 is achieved when p=10/7. The maximum of 7√101 is achieved when p=±10. (This is like going 40 right and 4 up followed by 30 units left and 3 up.)

[@ramakrishnaniyer889]

Extremely interesting problem. And what an elegant geometrical solution. So simple.

[@leif1075]

Miss, do you think really anyone would think of the triangle method..why not think of those as circles since those could be the equation of circles with these quantities being the radius?

[@justpaulo]

I saw this problem yesterday at night. I thought a bit about it, but went to bed w/o solving it. The similarity of the sqrts with hypotenuses of right triangles was striking, but I was imagining x, y and z as the 3D axes and couldn't wrap my head around it. Just now I decided to give it another shot, and when I was about to quit suddenly it occured me that x, y and z could be all in the number line , and from there to a straight line going from (0,0) to (10,7) is was a small jump. This was a fun hard problem and I'm happy that I succeeded, although I suspect I would not be able to do it under pressure in an AMC exam. PS: The geometrical solution is very elegant, but I really appreciated the calculus one b/c I had no idea how to approach it (until now). Thanks for that !

[@leif1075]

Isnt it kind of obvious it is minukum when each square root expression is minimized so just set all the variables equal to zero and you got your answer..1 plus 2 plus 1 plus square root of 90...that's the answer..

[@leif1075]

Hi2 is it hard? Cant you conclude each variable should wqual zero to minimize the values under the square root and then just sum the result 1 pmus square root of 4 plus 1 plus square root of 10 plus 9 and you're done..

[@justpaulo]

@@leif1075 Nope. x=y=z=0 do minimize the 1st three square roots, but on the 4th (on the right) you actually want to have some Z in order to minimize the (10-z)². Setting some Z ≠ 0 will actually "back propagate" into the other three square roots... Btw, the last square root is not √(10+9) like you say if z=0, but rather √(100+9) b/c the (10-z) is squared. With x=y=z=0 the end result would be 1+2+1+sqrt(109) = 14.44 which is bigger than the correct result of sqrt(149) = 12.2. Another (better) way to see it is to look at the geometrical interpretation at 6:42. If you make x=y=z=0 then you will not be drawing a straight line between the points (0,0) and (10,7). If you make all variables =0 you will actually draw a vertical line up to (0,4) followed by a line from (0,4) to (10,7). The actual x, y & z solutions to this problem are: x= 10/7 y= 30/7 z= 40/7

[@leif1075]

@@justpaulo but if you det z equal to nonzero, anythibf squared is positove and the sauare rootbof a larger number will necessarily be larger than that of a smaller number..and 10 minus z squared yes gives you the negative 20z term but also positive 3z and positive 100..so how does that not makebthe3value bigger?

[@justpaulo]

@@leif1075 The negative 20*z makes a big difference. For instance, if you make z=1, when compared to z=0, the z² term gives you a penalty of 1, but the -20z term gives you a bonus of -20. So z=1 makes the grand total smaller when compared to z=0. If you just google "graph (10-z)^2" you'll find out that it is a parabola which has a minimum at z=10. You can also try to convince yourself by calculating (10-z)² for several values of Z. Take for instance z=0 and z=10: z=0 => (10-z)² = 10² = 100 z=10 => (10-z)² = 0² = 0 This means that z=0 does not minimizes (10-z)² and therefore does not minimizes the sqrt( (10-z)² + 9). But z=10 does. However, if you make z=10 and keep y=0, now the 3rd term sqrt( (z-y)² + 1) is not =1 anymore, but rather =sqrt(101). Make y=10 to solve this problem and you just pushed the problem into the 2nd term. Make x=10 to minimize the 2nd term and now your 1st term becomes =sqrt(101).

[@Soaring-Dragon]

Assuming 0

[@Nevermind-sj6xu]

What a beautiful solution

[@arandomghost8819]

Multivariable calculus is always an option but isnt necessarily the best option....thx for providing us with an alternate approach as it may end up as a really handy tool... Edit :typo

[@satyanarayanmohanty3415]

Brilliant approach.

[@tthtlc]

your answer is astounding!!!!

[@prateeksaurabh]

waooo such a beautiful explanation. We humans tend to comprehend everything better visually. Thanks giving another perspective to how to think about maths problem. waooo

[@Rayhuntter]

hm, I rapidly and foolishly assumed it's 2.5 everywhere not taking into account that sqrt() is actually decreasing in growth rate with x increasing. It was interesting to note that in the end each 'segment' got a part of the '10-line' proportional to the sqrt() of its free coefficient. (1:2:1:3)

[@juliancarax4797]

well this was an interesting way of solving it

[@akashtutorials4865]

Wow superb explanation.. thank you

[@MuffinsAPlenty]

Great video!

[@utentecomputerman9941]

Fantastic

[@ДенисЛогвинов-з6е]

I tried to use that a+b> or = 2sqrt(ab) and this is = when a=b so I decided to get sqrt of both sides and then it was... |x|=1; |z-y|=1; |y-x|=4 and finally |10-z|=9 and this is hard to solve. Moreover the solution for x,y,z have to be awful because of real final answer...

[@amritlohia8240]

In the calculus solution, she obtains the values x = 10/7, y = 30/7, and z = 40/7. So they're not "awful".

[@poroutroangulo9606]

I'm madly in love whith...