Verified answer

The straightforward and easy method is using calculus. But there is a lot of algebra involved in this method. It's especially time consuming in a contest if no calculators are allowed.

Basically, taking first partials with respect to a and b and setting them equal to zero, we find a = 35 sqrt(2) / 17 and b = 35 / 12 is the only solution. Since the second partials at these a and b values are positive (and D = ∂_aa ∂_bb - (∂_ab)² > 0), this must be a minimum and f at that point is 13.

I can't think of any non-calculus way. Maybe it's useful to write your expression like:

√((a - 7/ √2)² + (7/√2)² ) + √((a - b/√2)² + (b/√2)² ) + √((b - 5)² + 5²)

********

Well, I don't know about the algebra being "killer" but there's just a lot of it. Let me give some more details what the algebra will be like.

f(a,b) = your expression

∂ f/∂ a = (2a - 7 √2) / (2√(49 - 7√2a + a²)) + (2 - √2b) / (2√(a² - √2ab + b²)) = 0 (we require this to be zero for a "critical point.")

Now bring one of the fractions to other side, square both sides and multiply out, you get alot of terms but many of them will cancel and what will be left is

2a²b² - 98a² + 98√2ab - 14√2ab² + 2a²b² = 0

which prettily factors out to

2a (b - 7) (7a +ab - 7√2b) = 0

but a ≠ 0 and b ≠ 7 because they don't make ∂_a zero.

So we can now cancel 2a (b - 7) and solve for a in 7a +ab - 7√2b = 0.

a = 7√2b / (b + 7).............................eq (1)

Here we are almost half way done in finding the answer.

∂ f/∂b = (2b - 10) / (2(√50 -10b + b²)) + (2b - √2a) / (2√(a² - √2ab + b²)) = 0 ..............eq(2)

again do the same as we did for ∂_a. A lot of terms will cancel and we get:

a²b² - 50b² - 10a²b + 50√2ab = 0

or

a²b - 50 b - 10a² + 50√2a = 0

Plug in what we have in eq(1) for a in above and simplify to get:

2b (24b² - 490b +1225) / (7 + b)² = 0

b ≠ 0 and (7+b) ≠ 0 so we are left with the quadratic

24b² - 490b +1225 = 0

Answers for b are 35 / 12 and 35 / 2. But the latter doesn't work when we plug it in eq(2). so the only solution is b = 35 / 12. Pluging this in eq(1) we get a = 35 √2 / 17.

The rest is just plugging in numbers. What can take a lot of time is taking second partials and evaluating them to prove it's a minimum (the numbers we have to work with get quite large - up to millions).

The biggest challange here is not to make a computational mistake. If this is a multiple choice contest, I'm definitely going for the estimation methods!

Here is my working for:

f(a,b) = √(49 + a² -7√(2) a) + √(a² + b² - √(2) ab) + √(50 + b² -10b)

[I put up the method first, but Remo did a much better job of linearizing it. I worked on the basis of JD's value of c≈2.9143, so if that was off, that explains why the rest of my algebra is off.

I'm curious how JD's and Remo's values of amin differ so much.]

Let's see can we write it partially or mostly involving quadratic forms:

f(a,b) = √((a -7/√2)² +(49-(7/√2)²)) + √((a-b)² +(2-√2)ab) + √((b-5)² +25)

= √((a -7/√2)² + 24.5) + √((a-b)² +(2-√2)ab) + √((b-5)² +25)

= p(a) + q(a,b) + r(b) ; say

[Remo pointed out my mistake, I had previously written:

f(a,b) = √((a -7/√2)² +49) + √((a-b)² +(2-√2)ab) + √((b-5)² +25)

so that slightly invalidates everything below this line.

===

Clearly the individual min values of p(a)=4.95 and r(b)=5. q(a,b) will be small when a≈b.

Note that q(a,b) can also be written as this quadratic form:

√ [ (1/2^(1/4))(a-b)² + (1- 1/√2)(a²+b²) ]

not sure that's useful though.

Let's try setting (a+b)/2 = c , (a-b)/2 = d. Thus a = c+d, b = c-d

We suspect d is very small, i.e. |c| >> |d|, thus |d/c| << 1

[According to JonDihon, c≈2.9143, d≈0.0025. Let us assume we found c≈2.9143 by calculus or interval-bisection.]

f(c,d) = √((c+d -7/√2)² +49) + √(4d² +(2-√2)(c+d)(c-d)) + √((c-d-5)² +25)

= √((c+d -7/√2)² +49) + √((6-√2)d² +(2-√2)c²) + √((c-d-5)² +25)

This is a very useful exact formulation.

We can get an approximation to f(c,d) by setting d≈0

f(c,0) = √((c -7/√2)² +49) + √((2-√2)c²) + √((c-5)² +25)

= √((c -7/√2)² +49) + √(2-√2) c + √((c-5)² +25)

= 7√(1 + ((c -7/√2)/7)² ) + √(2-√2) c + 5√(1 + ((c-5)/5)²)

I was thinking of Binomial expansion on that, but it's better to first note 7/√2≈4.95 and set C=c-5

f(C) ≈ 7√(1 + ((C - 0.05025)/7)² ) + √(2-√2)(C+5) + 5√(1 + (C/5)²)

≈ 7 [1 + ((C - 0.05025)/7)²/2] + √(2-√2) (C+5) + 5 [1 + (C/5)²/2] from first-order Binomial expansion

≈ [C² -C/10 +1/25 ] /14 + [12 +5√(2-√2)] + [√(2-√2)] C + [1/10] C²

≈ [12 +5√(2-√2) +1/350] + [√(2-√2) - 1/140] C + [1/14 + 1/10] C²

≈ 15.83 + 0.7582 C + 0.07 C²

Actually this particular approximation sucks since c wasn't that close to 5 to start with. But you get the approximation idea.

If you can pick a good approximation and then use the Binomial expansion to linearize it and find approximate cmin.

Then rewrite f(c,d) as f(cmin, d). Should allow you to find dmin.

Picking cmin≈2.9143 gives messy algebra.

So I tried backtracking to our exact formulation:

f(c,d) = √((c+d -7/√2)² +49) + √((6-√2)d² +(2-√2)c²) + √((c-d-5)² +25)

It's not much use to write d/c = x, then (c+d) = c(1+x), (c-d) = c(1-x)

So try minimizing f(c,0) by calculus:

f(c,0) = √((c -7/√2)² +49) + √((2-√2)c²) + √((c-5)² +25)

f'(c,0) = (c -7/√2)/ √((c -7/√2)² +49) +√(2-√2) + (c-5)/ √((c-5)² +25)

Setting C = c-5

f'(c,0) = (C +0.05)/ √((C +0.05)² +49) +√(2-√2) + C/ √(C² +25)

Iteratively we would find Cmin = cmin-5 = 2.9143-5 = -2.0857, so use that to simplify the denominators:

f'(c,0) = (C +0.05)/ 7.29 +√(2-√2) + C/ 5.4176

f'(c,0) = 0 =>

C/ 7.29 + C/ 5.4176 = -√(2-√2) -0.05/7.29

0.3218 C = -0.7722

C = -2.4 ; Umm, my algebra seems a bit off.

Anyway you can get cmin that way.

Then f(cmin,d) = √((c+d -7/√2)² +49) + √((6-√2)d² +(2-√2)c²) + √((c-d-5)² +25)

≈ √((2.9143+d -7/√2)² +49) + √((6-√2)d² +(2-√2)2.9143²) +(2-√2)2.9143²) + √((2.9143-d-5)² +25)

≈ √((d -2.0354)² +49) +

√(2-√2) *2.9143 [ 1 + (√(6-√2)/2*2.9143*√(2-√2)) d ]

+ √((2.0857+d)² +25)

≈ 7√(1 + ((d -2.0354)/7)²) + 5 √(1 + ((2.0857+d)/5)²)

+ 2.2305 [ 1 + 2.141/4.461 d ]

[now neglecting the quadratic terms (d/5 or 7)² ...]

≈ 7√( [1 +(2.0354/7)²] - 2(2.0354)d/7)

+ 5 √( [1 +(2.0857/5)²] +2((2.0857)d/5)

+ 2.2305 + 1.0707 d

≈ 7√(1.08455 - 0.5362d) + 5√(1.174 + 0.83428d) + 2.2305 + 1.0707 d

≈ 7.2899√(1 - 0.5362d) + 5.41756√(1 + 0.71063d) + 2.2305 + 1.0707 d

≈ 7.2899 [1 - 0.5362d/2 ] + 5.41756 [ 1 + 0.71063d/2] + 2.2305 + 1.0707 d

≈ 14.938 + 1.04122 d in the vicinity of c≈2.9143 and small d

I guess we needed to propagate quadratic terms to get dmin and thus accurate fmin.

My working seems to say fmin > 14.4 which is a little off.

As noted I worked on the basis of JD's value of c≈2.9143, so if that's off, everything else will be off. If I had the energy I would redo all this for Remo's cmin≈2.76

I'll be interested to see what methods other people suggest...

**Method 1:

Adding to SMCI:

f(a,b) = √((a -7/√2)² +49) + √((a-b)² +(2-√2)ab) + √((b-5)² +25)

Note that the money is on term √((a-b)² +(2-√2)ab)

The first term can be written as ~= 7 + .07(a -7/√2)^2

And the third term can be written ~= 5 + .1 (b-5)^2

Now, if you assume a~=b, then the middle term ends up being ~= .76* sqrt(ab) ~= .76a ~= .76b

Much easier.

********

While I was working, SMCI, added some stuff and is heading in the same direction as me. If you don't have a computer or a fancy calculator to help you out, this will become a time v. money question.

*******

let's go 1 more step:

f(a,b) ~= 12 + .07(a - 4.95)^2 + .1(b- 5)^2 + .76sqrt(ab)

Now, assume a +.05 = b

f(a,b) ~= 12 + .17(b- 5)^2 + .76b

f(a,b) ~= 12 + .17(b^2- 10b +25) + .76b

= .17b^2 -.94b + 16.25

Now diff to get min:

0= .34b -.94

b = 2.76

Note: I made some reasonable assumptions to get an approximation. I am not solving the function but trying to estimate the minimum

>> See correctlon below.

**********

My number is off. I see where I went off -- copying SMCI's first equation.

[Redacted, SMCI did the equation right -- I can't read -- need new glasses]

*********

SMCI -- I'm sorry, your equation is correct. Somehow when I read it a second time question a second time while trying to find my mistake, I read in a "/" sign which wasn't there.

....................

On further analysis, my number above "2.76" is almost right. It comes up with a min of 13.00399, and using JD's numbers, which is better, the min is 13.00027.

I used one significant digit to get my solution. Not bad all things considered. lol.

Given how flat the curve is, that is the best I can do. It is a simple first order approximation.

***********

**Method 2

Just looking at this one more time I am impressed by the symetry, You could just substitute a=b and the first term = third term and get

2√(50 + a² - 10a) + √(a² (2 - √(2))

take the derivative

0 = df/da =2 * (a-5)/sqrt(50 + a² - 10a) + .7653

.7653 sqrt(50 + a² - 10a) = 2 (a-5)

Now square everything

50 + a² - 10a = 6.8284 (a² -10a +25)

0= 5.8284a² -58.284a + 120.71

0= a² -10a + 20.7106

You get roots of 2.9289 and 7.0711

since you can reject the 7.07011 root the min would be 2.9289

Note: Again, I made some reasonable assumptions to get an approximation. This is more accurate.

*******

SCMI: JD's number is damn close. I ran out some numbers and got the same result. Here is my summary:

1. a= 2.9117, b= 2.9168, Min = 13.00026918 ......JD interpolating

1. a= 2.9110, b= 2.9160, Min = 13.00026918 ..... Remo interpolating

2. a= 2.9289, b= 2.9289, Min = 13.00030182 .... Remo method 2

3. a= 3.0480, b= 3.0480, Min = 13.00315426.... Remo method 1, quick and dirty -- corrected per below.

************

I knew something was wrong with term 1 in Method 1. I had

√((a -7/√2)² +49)

It should be

√((a -7/√2)² +(49- (7/√2)²)), or

√((a -4.95)² + 24.5)

~= 4.95 + .098(a -4.95)²

also,

f(a,b) ~= 9.95 + .196(b^2- 10b +25) + .7653b

f' = .392b - 1.195

b= 3.048

This does not improve much on the accuracy of Method 1. But it is the correct number per method.

SMCI: You may wish to note this because it well help you. It will knock ~2 off of your answer.

By an iterative process I get a minimum value of 13 when a ~ 2.9117 and b ~ 2.91679.

EDIT:

Me too.