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

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

Differentiating and equating it to zero we get


but its very tedious to find critical points here.

any other approach?

  • $\begingroup$ You haven't differentiated it correctly. $\endgroup$ – Shubham Johri Dec 6 '18 at 20:05
  • $\begingroup$ According to wolframalpha, the minimizing $x$ is very ugly: $x=\frac{300-3\sqrt{4441}}{763}$. The minimum value ain't so bad: $\sqrt{109}$. So, even if you did the differentiation correctly you would get some inconvenient quadratic to deal with. $\endgroup$ – user614671 Dec 6 '18 at 20:16

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.

  • $\begingroup$ excellent analysis $\endgroup$ – Umesh shankar Dec 8 '18 at 9:18


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

the function becomes

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

$F'(t)=0$ gives

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

  • $\begingroup$ Say: Yes, quick hint. Should be maybe a comment, not an answer. $\endgroup$ – dan_fulea Dec 6 '18 at 20:06
  • $\begingroup$ And what about then? $\endgroup$ – Aqua Dec 6 '18 at 20:11
  • $\begingroup$ @greedoid Diffetentiate with respect to $t$. it is easier. $\endgroup$ – hamam_Abdallah Dec 6 '18 at 20:11
  • 1
    $\begingroup$ @greendoid well, we only need the points, not the point of the idea. $\endgroup$ – dan_fulea Dec 6 '18 at 20:12
  • $\begingroup$ @hamam_Abdallah then it is also simple to type this differential and solve w.r.t. $t$. Just fill in the details please. There is a difference between a possible beginning and and answer. $\endgroup$ – dan_fulea Dec 6 '18 at 20:13

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 $x\in J$ $$ \begin{aligned} A(x) &= \sqrt{58-42 x}\ ,\\ B(x) &= \sqrt{149-140\sqrt{ 1-x^2}}\ .\\ &\qquad\text{Then we have }\\ 10^2(58-A^2)^2 + 3^2(149-B^2)^2 &= 420^2\ . \end{aligned} $$ So we can formulate an equivalent problem:

Minimize $a+b$ constrained to $a$ between $\sqrt{58\pm 42}$ (i.e. $3$ and $10$), and $b$ between $\sqrt{149\pm 140}$ (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:

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()


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:
  *(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
    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: $$ a_* = \sqrt{\frac1{109}(4522+18\sqrt{4441})}\ . $$ And the corresponding $x_*=(58-{a_*}^2)/42$ is $$x_* = \frac 3{763}(100-\sqrt{4441})\approx 0.13116319137804\dots\ .$$

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] )

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