Let $a,b,c \in \mathbb{R}^{+}$ and $f(x)=\sqrt{a^2+x^2}+\sqrt{(b-x)^2+c^2}$. Then $\min f={?}$

Question

Let $a,b,c \in \mathbb{R}^{+}$ and $f(x)=\sqrt{a^2+x^2}+\sqrt{(b-x)^2+c^2}$. Then $\min f={?}$

My Try : $f'(x)=\dfrac{x}{\sqrt{a^2+x^2}}+\dfrac{(x-b)}{\sqrt{(b-x)^2+c^2}}=0$

$x^2((b-x)^2+c^2)=(x-b)^2(a^2+x^2) \ \ \to x^2(x-b)^2+x^2c^2=(x-b)^2a^2+x^2(x-b)^2$

$x^2c^2=(x-b)^2a^2 \to x^2c^2=(x^2-2bx+b^2)a^2 \to x^2c^2=x^2a^2-2ba^2x+b^2a^2$

$(c^2-a^2)x^2+2ba^2x-b^2a^2=0 \to x=\dfrac{-ba^2\pm \sqrt{ba^2(c^2-a^2)(-b^2a^2)}}{c^2-a^2}$

Now what ? and also It can be obtained without the use of a derivative?

Hint: try to interpret this function geometrically, specifically as a sum of distances. — Paco Adajar, Aug 16 '20 at 07:35
See https://math.stackexchange.com/questions/3791835/if-x-is-a-nonnegative-real-number-find-the-minimum-value-of-sqrtx2-4/3791860#3791860 — A. Goodier, Aug 16 '20 at 07:40

score 6 · Accepted Answer · answered Aug 16 '20 at 07:44

6

See the picture, using triangle inequality, we have

$\sqrt{a^{2}+x^{2}}+\sqrt{(b-x)^{2}+c^{2}}\geq\sqrt{(a+c)^{2}+b^{2}}$

answered Aug 16 '20 at 07:44

acat3

very nice thanx, now what is Minimum point length (x) =? – justin32 Aug 16 '20 at 08:05
1

@justin32 we get minimum if our "triangle" is actually a line. Therefore $\frac{a}{x}=\frac{a+c}{b}$ – acat3 Aug 16 '20 at 08:07
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@justin32 Please consider accepting the answer if you found it helpful. – cr001 Aug 16 '20 at 09:42

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