Encuentra el mínimo de <span class="math-container">$$f(x,y)=\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$$ It seems that <span class="math-container">$f_x=f_y=0$</span> es muy difícil de calcular. ¿Hay alguna idea más fácil?</span>
Respuestas
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guest
Puntos
1
Completa el cuadrado en los dos primeros términos y deja que <span class="math-container">$Y=2y-3$</span> y <span class="math-container">$X=3x-3/2$</span> obtengan <span class="math-container">$$f(X,Y)=\sqrt{Y^2+1}+\sqrt{2X^2+\frac12}+\sqrt{2\left(X+\frac12\right)^2+\left(Y+\frac12\right)^2-2XY+\frac74}$$ so that <span class="math-container">\begin{align}f_X&=\frac{2X}{\sqrt{2X^2+1/2}}+\frac{2X-Y+1}{\sqrt{2(X+1/2)^2+(Y+1/2)^2-2XY+7/4}}=0\f_Y&=\frac Y{\sqrt{Y^2+1}}+\frac{Y-X+1/2}{\sqrt{2(X+1/2)^2+(Y+1/2)^2-2XY+7/4}}=0\end{align}</span> which can be rewritten as <span class="math-container">\begin{align}\frac{4X^2}{2X^2+1/2}&=\frac{(2X-Y+1)^2}{2(X+1/2)^2+(Y+1/2)^2-2XY+7/4}\\frac{Y^2}{Y^2+1}&=\frac{(Y-X+1/2)^2}{2(X+1/2)^2+(Y+1/2)^2-2XY+7/4}\end{align}</span> and extensively simplified to give the symmetrical forms <span class="math-container">\begin{align}X^2(4Y^2+16Y+12)+4(Y-1)X-(Y-1)^2&=0\Y^2(4X^2+12X+5)+4(2X-1)Y-(2X-1)^2&=0\end{align}</span> on clearing denominators. Equating the two expressions on the left gives <span class="math-container">$%(4X-3Y)(2XY+2X+Y-1)=0$8X^2Y-6XY^2+8X^2-3Y^2-2XY-4X+3Y=0$X=3Y/4$</span> which factorises to <span class="math-container">$Y=-2,\pm1/3$</span>, where the two interaction terms significantly aid this observation. Substituting <span class="math-container">$X=-3/2,\pm1/4$</span> gives the polynomial <span class="math-container">$\sqrt{10}$9Y^4+36Y^3+35Y^2-4Y-4=0$%%%2XY+2X+Y-1=0$</span> which has roots at <span class="math-container">$f(X,Y)$</span> after using the rational root theorem. Hence <span class="math-container">$f_X=0$</span> and a minimum of <span class="math-container">$X\le-3/2$</span> is obtained at <span class="math-container">$(X,Y)=(0,1)$(X,Y)=\left(-\frac14,-\frac13\right)\implies(x,y)=\left(\frac5{12},\frac43\right).$f_Y=0$</span> In fact, it is the global minimum as the expression <span class="math-container">$-3/2\le X is an anomaly due to squaring. The square root in the third term of <span class="math-container">$(X,Y)=(1/2,0)$</span>.</span></span>
Jared
Puntos
8
Tenemos <span class="math-container">$$f\left ( x, y \right ):= \sqrt{4y^{2}- 12y+ 10}+ \sqrt{18x^{2}- 18x+ 5}+ \sqrt{18x^{2}+ 4y^{2}- 12xy+ 6x- 4y+ 1}:=$$</span> <span class="math-container">$$\sqrt{\frac{\left ( 2y- 6 \right )^{2}+ \left ( 6y- 8 \right )^{2}}{10}}!+!\sqrt{\frac{\left ( 6x- 5 \right )^{2}+ \left ( 12x- 5 \right )^{2}}{10}}!+!\sqrt{\frac{\left ( 6x+ 2y- 1 \right )^{2}+ \left ( 12x- 6y+ 3 \right )^{2}}{10}}$$</span> <span class="math-container">$$\geq\frac{6- 2y}{\sqrt{10}}+ \frac{5- 6x}{\sqrt{10}}+ \frac{6x+ 2y- 1}{\sqrt{10}}= \sqrt{10}$$</span>
