Question

find the maximum value of \[25(a-b)^2 + (\sqrt{100-4a^2}+5\sqrt{1-b^2})^2\]

Answer 1

Okay so I see some logic over here. The previous question that I asked was something similar to the expression, but we are required to find the maximum value here. Not a big deal. Let us try it out.

Let \(\sqrt{100-4a^2} = y\)

\(y^2 + 4a^2 = 100\)
\(y^2 + (2a)^2 = 100\)

Seems more like a circle with center \((y,2a)\) and radius \(10\) units.

Now, \(5\sqrt{1-b^2} = x\)
\(x^2 = 25(1 - b^2)\)
\(x^2 = 25 - 25b^2\)
\(x^2 + 25b^2 = 25\)
\((x)^2 + (5b)^2 = 5^2\)

o.O So we have got another equation for a circle, with center : \((x,5b)\) and radius \(5\) units.

Answer 2

thats an excellent approach @mathslover :)
you're just 2 steps away from picking the right conics that work !

Answer 3

So, I took right conics or not?

Answer 4

looks like you almost have them! but they still require some tweaking..
especially the 25 that is attached to the first term in the expression requires special care :P

Answer 5

Hmm! :/

\((5a - 5b)^2 + (\sqrt{100-4a^2} + \sqrt{25(1-b^2)} )^2 \)

Let \(y_1 = \sqrt{100 - 4a^2}\) \(\implies {y_1}^2 = 100 - 4a^2\)

\({y_1}^2 + 4a^2 = 100 \)

Let \(y_2 = \sqrt{5(1-b^2)}\) \(\implies {y_2}^2 = 25(1-b^2) \)

\({y_2}^2 + 25b^2 = 25 \)

Hmm! :/

Answer 6

Am still trying... :(

Answer 7

That "+" in the second term is annoying...
\(25(a-b)^2 + (\sqrt{100-4a^2} \color{red}{+} 5\sqrt{1-b^2})^2\)

Answer 8

\[25\left[(a-b)^2 + \left(\dfrac{\sqrt{100-4a^2}}{5}+\sqrt{1-b^2}\right)^2\right]\]

Answer 9

spotting the conics might be simple now ?

Answer 10

My head is still spinning from this this morning. I'm gonna sit this out and watch.

Answer 11

Cool! Let's see what's the difference:

\((a-b)^2 + \left( \cfrac{\sqrt{100-4a^2} }{5} + \sqrt{1-b^2} \right)^2 \)

I can take two obvious points from here, the x coordinates of both the points will be : \(a\) and \(b\)

Let us now try to focus upon y -coordinates.

Let \(\cfrac{\sqrt{100-4a^2}}{5} = y_1\)

\(25{y_1}^2 = 100 - 4a^2\)
\(25{y_1}^2 + 4a^2 = 100\)

Cool! We have, a circle with center \((5 y_1 , 2a)\) and radius = 10

Next :

Let \(y_2 = \sqrt{1-b^2} \\
{y_2}^2 = 1 - b^2 \\
{y_2}^2 + b^2 = 1 \)

So, we have a circle with center \( (y_2 , b) \) and radius = 1

We've \(\left(a ,\cfrac{\sqrt{100-4a^2}}{5} \right) \) and \(\left(b , \sqrt{1-b^2} \right)\)

So, we are trying to find maximum distance between two circles of given centers as \((5 y_1 , 2a)\) and \( (y_2 , b) \) respectively.

Answer 12

Weird. I'm pretty sure that I'm doing something wrong. I will check everything again, one sec.

Answer 13

No ,you're doing great :) just this :

\(25{y_1}^2 + 4a^2 = 100 \) is more like an ellipse... replacing \(a\) by \(x\) makes it easy to see :

\[25y^2 + 4x^2 = 100 \]

\[\dfrac{y^2}{4} + \dfrac{x^2}{25} = 1\]

Answer 14

o.O :O O.o Didn't observe that ... nice..! So, we have one circle and one ellipse.. hmm!

Answer 15

the other conic is an unit circle as you guessed, but it is centered at origin !

Answer 16

My concepts... :/ I need to go back and revise my concepts. :(

Well, its really a nice question. I will try to proceed from here.

Answer 17

Okay... :) im not giving any useful hints as i don't want to spoil the fun for you !

Answer 18

hmm then what just maximize
x^2/25+y^2/4=1
with partial derivatives ?

Answer 19

how is y^2+4a^2=100 a circle?

Answer 20

it is an ellipse @xapproachesinfinity

Answer 21

25y^2+4a^2=100

y^2/4 + a^2/25 = 1

Answer 22

he got me lost when he said it is a circle lol

Answer 23

yeah i recognize the ellipse equation there

Answer 24

:) the question is still open, mathslover solved like 90% of it but its gona take a while for others to understand what he was doing...

Answer 25

i'm not really acquainted with ellipses and hyperbolas
i need to review their equations
=============
can we use calculus here?

Answer 26

any method is fine :) i feel calculus might give nasty derivatives for solving hmm

Answer 27

seeing the given expression as related to the `distance formula` is the key here, but im really looking forward to find ANY other cool ways to approach this problem xD

Answer 28

i see then i'll work on it :)

Answer 29

I got it.

Answer 30

Is it 15? @ganeshie8

Answer 31

ok here is one method ( reducing error to get high accurate )
|a|<=5
|b|<=1
and we need |a-b| br the largest difference , i'll stop here so other reveal more excited solution , looking forward for the geometrically solution sounds interesting !

Answer 32

nice :) 15 for distance or the max value ?
http://www.wolframalpha.com/input/?i=maximize+25%28a-b%29%5E2+%2B+%28sqrt%28100-4a%5E2%29%2B5sqrt%281-b%5E2%29%29%5E2

Answer 33

For the maximum distance ....

Answer 34

You have correct numbers for bounds @Marki ! xD

Answer 35

;) also solution

Answer 36

one sec mathslover, i only saved the maximum value of given expression..

Answer 37

Okay so here is what I've done:

I've point as : \( \left(5a , \sqrt{100 -4a^2} \right) \) and \(\left(5b, -5\sqrt{1-b^2} \right) \)

2 Conics that we have are :

\({y_1}^2 + 4a^2 = 100 \\
\cfrac{{y_1}^2}{100} + \cfrac{a^2}{25} = 1 \ \ \ \ \ \ \ \ \ [A]\)
a = 10 and b = 5

And \(\cfrac{{y_2}^2}{25} + \cfrac{b^2}{1} = 1 \ \ \ [B]\)
a = 5 and b = 1

So, we have two equations \([A]\) and \([B]\) for both ellipse. We need to find maximum distance between these ellipse which will be in fact sum of semi-major axis of larger ellipse (A) and semi-major axis of (B)

So, we have maximum distance as : 10 + 5 = 15 units

Answer 38

Excellent! but there are few numbers misplaced here and there - dont wry about it :) method is more important, not so much the final answer !

Answer 39

I will check them too! :)

Answer 40

One sec.

Answer 41

I've been waiting for the answer to this for a couple days now, I can't wait, this is exciting! =D

Answer 42

\[25\left[(a-b)^2 + \left(\dfrac{\sqrt{100-4a^2}}{5}+\sqrt{1-b^2}\right)^2\right]\]

the stuff inside square brackets can be interpreted as the square of distance between below conics :

\(y = \dfrac{\sqrt{100-4x^2}}{5}\)

\(y = \sqrt{1-x^2}\)

Look at the graph and eyeball the maximum value between them :
http://gyazo.com/572956ac53a4dfb4a2884a451e28bf2b

the maximum distance is 5--1 = 6
square it and multiply by 25 for the maximum value of the required expression

Answer 43

Aw, can you help me in identifying the mistakes in my method?

Answer 44

there is one loophole in my solution though :P

Answer 45

@mathslover factoring out 25 will give you a circle and ellipse
carrying out your exact same method yields the same answer given by wolfram :)

Answer 46

i don't see anything wrong in your steps..

Answer 47

Yeah :/ As factoring out 25 will not make any differences in the final answer, isn't it?

And from my answer, I've the maximum value as (15)^2 = 225... does this match to your final answer?

Answer 48

max has to be 900
http://www.wolframalpha.com/input/?i=maximize+25%28a-b%29%5E2+%2B+%28sqrt%28100-4a%5E2%29%2B5sqrt%281-b%5E2%29%29%5E2

Answer 49

This seems like the hard way to do Lagrange multipliers lol.

Answer 50

Cool! We just need to add 675 to my answer and you're done... LOL Not a big deal ... :P

Answer 51

well as i got same to ganesh approch
|a-b|=6
so
25(6^2)=900

Answer 52

Absolutely! not really a big deal at all haha ;) it would be more fun to find the minimum distance between a parabola and circle as eyeballing wont work in that case

Answer 53

I totally agree =)