# Find the minimum of |a| - |b|

#### anemone

##### MHB POTW Director
Staff member
If $\log_4 (a+2b)+\log_4 (a-2b)=1$, find the minimum of $|a|-|b|$.

#### chisigma

##### Well-known member
Re: Find the minimum of |a|-|b|

If $\log_4 (a+2b)+\log_4 (a-2b)=1$, find the minimum of $|a|-|b|$.
From the initial conditions we derive immediately...

$\displaystyle (a + 2\ b)\ (a - 2\ b) = 4 -> b = \frac{\sqrt{a^{2} - 4}}{2}\ (1)$

... so that the problem is to minimize respect to a the function...

$\displaystyle f(a) = a - \frac{\sqrt{a^{2} - 4}}{2}\ (2)$

Kind regards

$\chi$ $\sigma$

#### anemone

##### MHB POTW Director
Staff member
Re: Find the minimum of |a|-|b|

From the initial conditions we derive immediately...

$\displaystyle (a + 2\ b)\ (a - 2\ b) = 4 -> b = \frac{\sqrt{a^{2} - 4}}{2}\ (1)$

... so that the problem is to minimize respect to a the function...

$\displaystyle f(a) = a - \frac{\sqrt{a^{2} - 4}}{2}\ (2)$

Kind regards

$\chi$ $\sigma$
Thanks for participating, chisigma! I noticed you stopped half-way and probably you could eyeball the answer from where you have stopped?

Yes, your interpretation to the problem is correct and using the calculus method to find the minimum point of the function $\displaystyle f(a) = a - \frac{\sqrt{a^{2} - 4}}{2}\ (2)$, one will get the minimum value of $|a|-|b|=\dfrac{\sqrt{15}}{2}$ when $a=\dfrac{8}{\sqrt{15}}$.