Jtdylktuy

Find Minimum value of $$f(x)=sqrt{58-42x}+sqrt{149-140sqrt{1-x^2}}$$

My try: the domain of the function is $x in [-1 ,,,1]$

Differentiating and equating it to zero we get

$$f'(x)=frac{-21}{sqrt{58-42x}}+frac{70}{sqrt{1-x^2}sqrt{149-140sqrt{1-x^2}}}=0$$

but its very tedious to find critical points here.

any other approach?

Let $(x,y)$ be a point on the unit circle $x^2+y^2=1$. We have to minimize the function:
$$
begin{aligned}
f(x,y)
&=
sqrt{(7x-3)^2+(7y-0)^2} + sqrt{(7x-0)^2+(7y-10)^2}
\
&=
operatorname{Distance}( (7x,7y) , (3,0) )
+
operatorname{Distance}( (7x,7y) , (0,10) )
\
&ge
operatorname{Distance}( (3,0) , (0,10) )
=sqrt{3^2+10^2} ,
end{aligned}
$$
with equality in the $ge$ above only for the point of intersection of the segment with the above distance with the circle of radius $7$ centered in the origin.

hint

As $xin[-1,1]$, you can put
$$x=cos(t)$$ with $$0le tle pi.$$

the function becomes

$$F(t)=$$
$$sqrt{58-42cos(t)}+sqrt{149-140sin(t)}$$
$$frac 17F'(t)=$$
$$frac{3sin(t)}{sqrt{58-42cos(t)}}-frac{10cos(t)}{sqrt{149-140sin(t)}}$$

$F'(t)=0$ gives

$$9sin^2(t)(149-140sin(t))=$$
$$100cos^2(t)(58-42cos(t))$$

If you are not interested in the analytical zero of $f'(x)$ you could try to find the solution using e.g. the Bisection method, or a Fixed-point iteration, e.g. Newton's method. However, if you are going to be solving it by hand, I admit that this doesn't really help with the tedious part – in fact, it is probably worse.

Edit: Also, as noted in a comment, your differentiation is not correct. Check your second term.

In the following, $x$ runs in the interval $J=[-1,1]$.
We introduce the functions of $xin J$
$$
begin{aligned}
A(x) &= sqrt{58-42 x} ,\
B(x) &= sqrt{149-140sqrt{ 1-x^2}} .\
&qquadtext{Then we have }\
10^2(58-A^2)^2 + 3^2(149-B^2)^2 &= 420^2 .
end{aligned}
$$
So we can formulate an equivalent problem:

Minimize $a+b$ constrained to $a$ between $sqrt{58pm 42}$ (i.e. $3$ and $10$), and $b$ between $sqrt{149pm 140}$ (i.e. $3$ and $17$) and $$10^2(58-a^2)^2+3^2(149-b^2)^2= 420^2 .$$

So we search Lagrange multiplicators for the function
$$F(a,b;t)=(a+b)-t(10^2(58-a^2)^2+3^2(149-b^2)^2-420^2)$$
to get the local extremal points. (Then we still have to compare with the marginal values.)
We obtain the following system:
$$
left{
begin{aligned}
0 &= F'_a(a,b;t) = 1+10^2;4at;(58-a^2) ,\
0 &= F'_b(a,b;t) = 1+3^2;4bt;(149-b^2) ,\
0 &= F'_t(a,b;t) = 10^2(58-a^2)^2+3^2(149-b^2)^2- 420^2 .
end{aligned}
right.
$$
Algebraically, we have only a slight improvement, compared with the equation $f'(x)=0$, which was the intention in the OP, where we have some radicals. Above we have a purely algebraic system, the slight improvement, and still have to start the solution.

The idea is elimination.

We first eliminate $4t$, which appears linearly, getting:
$$
left{
begin{aligned}
10^2;a;(58-a^2) &= 3^2;b;(149-b^2) ,\
10^2(58-a^2)^2 &+3^2(149-b^2)^2= 420^2 .
end{aligned}
right.
$$
One possible elimination idea (of $b$) from this point is as follows.
We are squaring the first equation, so we have an expression of $b^2(149-b^2)^2$ in terms of $a$, so $(149-b^2)^2$ is a polynomial in $a$ divided by $b^2$. We insert this expression of $(149-b^2)^2$ in the second equation, thus obtaining a formula for $b^2$ as a polynomial in $a$. We insert this $b^2$ in the second equation, thus obtaining an equation only in $a$. Of course, i cannot stop here and say "from here it simple, details left to the reader"... This bloody job is explicitly as follows, using a computer,
sage in my case:

var('a,b,bb');

EXPR = ( 10^2*a*(58-a^2) / 3^2 )^2 / bb     # (149-b^2)^2

# bb is above a new variable for b^2

eq = solve( 10^2*(58-a^2)^2 + 3^2*EXPR == 420^2, bb )[0]

print "b^2 is the solution bb of:n%s" % eq 

bb = eq.rhs()

a_poly = 10^2*(58-a^2)^2 + 3^2*(149-bb)^2 - 420^2

print "a is a zero point for the expression:"

print a_poly.factor()

Results:

b^2 is the solution bb of:

bb == -100/9*(a^6 - 116*a^4 + 3364*a^2)/(a^4 - 116*a^2 + 1600)

a is a zero point for the expression:

  1/9

  *(100*a^8 - 13400*a^6 + 398509*a^4 + 2317356*a^2 + 47534400)

  *(109*a^4 - 9044*a^2 + 174400)

  /((a + 10)^2*(a + 4)^2*(a - 4)^2*(a - 10)^2)

(The last expression was manually broken to fit in page.)
So $a$ is a root of the one or the other polynomial in the numerator. So $a^2$ is either the root of $100 U^4 - 13400U^3 + 398509 U^2 + 2317356 U + 47534400$, or the root or $109 U^2 - 9044 U + 174400$.

I did the above "in a human manner", and against my taste and conviction.
Let us put it an other way.
Of course, we cannot expect to have "simple solutions" and a "simple elimination by quick hint", so let us the computers do the work for us, then we can still decide...

Using sage, we eliminate blindly:

sage: R.<a,b,t> = PolynomialRing(QQ)

sage: R

Multivariate Polynomial Ring in a, b, t over Rational Field



sage: F = a+b - t*( (10*(58-a^2))^2 + (3*(149-b^2))^2 - 420^2 )

sage: J = R.ideal( diff(F,a), diff(F,b), diff(F,t) )

sage: K = J.elimination_ideal([b,t])

sage: K

Ideal (10900*a^12 - 2365000*a^10 + 182067081*a^8 - 5688483592*a^6 + 53723051536*a^4 - 25754227200*a^2 + 8289999360000)

of Multivariate Polynomial Ring in a, b, t over Rational Field



sage: K.gens()[0].factor()

    (109*a^4 - 9044*a^2 + 174400) 

  * (100*a^8 - 13400*a^6 + 398509*a^4 + 2317356*a^2 + 47534400)

and this rather reflects my way to work.
We have thus all possible points / all candidates $a=sqrt{58-42x}$, so that calculating the corresponing $x$ (or the corresponding $b$) and inserting in $f$ (or $F$) can lead to a local minimum.

Let us finally ask for the numerical values, to finish, and get an answer. For each root $a$ of the above polynomial we compute numerically (to a good precision) the value of $f$. (Since the OP may not be interested in $F$.)

f(x) = sqrt(58-42*x) + sqrt(149-140*sqrt(1-x^2))

R.<a> = PolynomialRing(QQ)

P = (109*a^4 - 9044*a^2 + 174400) 

  * (100*a^8 - 13400*a^6 + 398509*a^4 + 2317356*a^2 + 47534400)



for aroot in P.roots(ring=AA, multiplicities=False):

    if aroot < 3 or aroot > 10:

        print "a = %s :: REJECTED" % aroot

        continue

    x = (58-aroot^2)/42

    print "a = %s x = %s f(x) = %s" % (aroot, x, QQbar(f(x)))

We get:

a = -8.857786578527434? :: REJECTED

a = -7.936142667572221? :: REJECTED

a = -7.245077360672018? :: REJECTED

a = -5.520990047273946? :: REJECTED

a = 5.520990047273946? x = 0.6552064023310009? f(x) = 12.09647575142790?

a = 7.245077360672018? x = 0.1311631913780425? f(x) = 10.44030650891055?

a = 7.936142667572221? x = -0.11862762952524599? f(x) = 11.096611974467387?

a = 8.857786578527434? x = -0.4871519778747795? f(x) = 14.02843358892943?

The minimal value of $f(x)$ among the above is obtained in the line:

a = 7.245077360672018? x = 0.1311631913780425? f(x) = 10.44030650891055?

(The value $f(x)$ is smaller than $f(pm1)$ and $f(0)$, corresponding to possible global minimal values at the boundary.)

The corresponding $a$ is a root of $109a^4 - 9044a^2 + 174400=0$, explicitly:
$$
a_* = sqrt{frac1{109}(4522+18sqrt{4441})} .
$$
And the corresponding $x_*=(58-{a_*}^2)/42$ is
$$x_* = frac 3{763}(100-sqrt{4441})approx 0.13116319137804dots .$$

Note: The numerical computations in between were done with a good enough precision to insure we pick the right candidate. So we have a proof. The following (truly) numerical computation is a check.

sage: var('x');

sage: minimize( sqrt(58-42*x) + sqrt( 149-140*sqrt(1-x^2) ), [0.5] )

(0.13116313434376808)