# Find Minimum value of $\sqrt{58-42x}+\sqrt{149-140\sqrt{1-x^2}}$

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

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

but its very tedious to find critical points here.

any other approach?

• You haven't differentiated it correctly. – Shubham Johri Dec 6 '18 at 20:05
• 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. – 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.

• excellent analysis – Umesh shankar Dec 8 '18 at 9:18

hint

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

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

• Say: Yes, quick hint. Should be maybe a comment, not an answer. – dan_fulea Dec 6 '18 at 20:06
• And what about then? – Aqua Dec 6 '18 at 20:11
• @greedoid Diffetentiate with respect to $t$. it is easier. – hamam_Abdallah Dec 6 '18 at 20:11
• @greendoid well, we only need the points, not the point of the idea. – dan_fulea Dec 6 '18 at 20:12
• @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. – 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:

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: $$a_* = \sqrt{\frac1{109}(4522+18\sqrt{4441})}\ .$$ 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] )
(0.13116313434376808)