0
$\begingroup$

For $a=\sqrt{x^2-3\sqrt2x+9}$ and $b=\sqrt{x^2-5\sqrt2x+25}$ what is the value of $x$ when $a+b$ is minimum and how to find this? Thanks in advance.

$\endgroup$
  • 1
    $\begingroup$ What have you tried? $\endgroup$ – Parcly Taxel Jan 28 '19 at 9:42
  • $\begingroup$ I tried to differentiate it with respect to x $\endgroup$ – Kshitij Singh Jan 28 '19 at 9:43
  • $\begingroup$ I tried to differentiate it with respect to x $\endgroup$ – Kshitij Singh Jan 28 '19 at 9:43
  • 1
    $\begingroup$ Please typeset your equations using Mathjax for better presentation $\endgroup$ – Shubham Johri Jan 28 '19 at 9:43
5
$\begingroup$

Let $\measuredangle ACB=90^{\circ},$ $AC=3$, $BC=5$ and $CD$ be a bisector of $\angle ACB$.

Also, let $CD=x$.

Thus, by the triangle inequality $$AD+BD\geq AB,$$ which gives $$a+b=\sqrt{x^2+3^2-2x\cdot3\cdot\cos45^{\circ}}+\sqrt{x^2+5^2-2x\cdot5\cdot\cos45^{\circ}}\geq\sqrt{3^2+5^2}=\sqrt{34}.$$ The equality occurs, when $D\in AB$, which says that we got a minimal value.

Now, by similarity we can show that $$CD^2=AC\cdot BC-AD\cdot BD$$ and since $$\frac{AD}{BD}=\frac{AC}{BC}=\frac{3}{5},$$ we obtain $$AD=\frac{3}{8}\sqrt{34},$$ $$BD=\frac{5}{8}\sqrt{34}$$ and $$x=\sqrt{3\cdot5-\frac{3}{8}\sqrt{34}\cdot\frac{5}{8}\sqrt{34}}=\frac{15}{4\sqrt2}.$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.