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Find the Minimum Value of |x|-|y|

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MHB POTW Director
Staff member
Feb 14, 2012
If \(\displaystyle \log_4(x+2y)+\log_4(x-2y)=1\), find the minimum value of $|x|-|y|$.


Staff member
Feb 24, 2012
My solution:

We are given the objective function:

\(\displaystyle f(x,y)=|x|-|y|\)

subject to the constraint:

\(\displaystyle \log_4(x+2y)+\log_4(x-2y)=1\)

which can be written as:

\(\displaystyle g(x,y)=x^2-4y^2-4=0\)

Using Lagrange multipliers, we obtain:

\(\displaystyle \frac{x}{|x|}=\lambda(2x)\)

\(\displaystyle -\frac{y}{|y|}=\lambda(-8y)\)

Observing that the critical point (0,0) is outside the domain of the constraint, we are left with:

\(\displaystyle \lambda=\frac{1}{2|x|}=\frac{1}{8|y|}\implies|x|=4|y|\implies x^2=16y^2\)

Substituting this into the constraint, we find:

\(\displaystyle 16y^2-4y^2=4\)

\(\displaystyle y^2=\frac{1}{3}\)

\(\displaystyle y=\pm\frac{1}{\sqrt{3}}\,\therefore\,x=\pm\frac{4}{\sqrt{3}}\)

Testing other point satisfying the constraint shows we have a minimum, hence:

\(\displaystyle f_{\min}=f\left(\pm\frac{4}{\sqrt{3}},\pm\frac{1}{\sqrt{3}} \right)=\frac{4}{\sqrt{3}}-\frac{1}{\sqrt{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}\)


MHB Oldtimer
Staff member
Feb 7, 2012
Geometric solution:

We want to find the closest that the hyperbola $x^2-4y^2=4$ comes to one of the lines $y = \pm x$. This will occur at points on the hyperbola where the gradient is $\pm 1$ (as in the green line in the picture above). Differentiate the hyperbola equation to get $2x - 8yy' = 0$, and put $y'=\pm1$ to find that $x=\pm4y$. Substitute that into the hyperbola equation, getting $12y^2=4$, or $y = \pm1/\sqrt3$, $x = \pm4/\sqrt3$. Thus $|x| - |y| = 3/\sqrt3 = \sqrt3$.