Respuesta :
Using Lagrange multipliers, we have the Lagrangian
[tex]L(x,y,z,\lambda_1,\lambda_2)=yz+xy+\lambda_1(xy-1)+\lambda_2(y^2+z^2-9)[/tex]
with partial derivatives (set equal to 0)
[tex]L_x=y+\lambda_1y=y(1+\lambda_1)=0[/tex]
[tex]L_y=z+x+\lambda_1x+2\lambda_2y=0[/tex]
[tex]L_z=y+2\lambda_2z=0[/tex]
[tex]L_{\lambda_1}=xy-1=0[/tex]
[tex]L_{\lambda_2}=y^2+z^2-9=0[/tex]
[tex]xL_x=xy(1+\lambda_1)=1+\lambda_1=0\implies\lambda_1=-1[/tex]
[tex]\implies L_y=z+2\lambda_2y=0[/tex]
[tex]zL_y=z^2+2\lambda_2yz=0[/tex]
[tex]yL_z=y^2+2\lambda_2yz=0[/tex]
[tex]\implies zL_y+yL_z=9+4\lambda_2yz=0\implies\lambda_2yz=-\dfrac94[/tex]
[tex]\implies zL_y-yL_z=z^2-y^2=0\implies z^2=y^2\implies y=\pm z[/tex]
Now,
[tex]\lambda_2yz=-\dfrac94\impliesyL_z=y^2-\dfrac92=0\implies y=\pm\dfrac3{\sqrt2}\implies z=\mp\dfrac3{\sqrt2}[/tex]
[tex]y=\pm\dfrac3{\sqrt2}\implies x=\mp\dfrac{\sqrt2}3[/tex]
So we have four critical points to look into: [tex]\left(\dfrac2{\sqrt3},\dfrac3{\sqrt2},\dfrac3{\sqrt2}\right)[/tex], [tex]\left(-\dfrac2{\sqrt3},-\dfrac3{\sqrt2},-\dfrac3{\sqrt2}\right)[/tex], [tex]\left(\dfrac2{\sqrt3},\dfrac3{\sqrt2},-\dfrac3{\sqrt2}\right)[/tex], and [tex]\left(-\dfrac2{\sqrt3},-\dfrac3{\sqrt2},\dfrac3{\sqrt2}\right)[/tex].
Respectively, these give extreme values of [tex]\dfrac{11}2[/tex] (max), [tex]\dfrac{11}2[/tex] (max), [tex]-\dfrac72[/tex] (min), and [tex]-\dfrac72[/tex] (min).
[tex]L(x,y,z,\lambda_1,\lambda_2)=yz+xy+\lambda_1(xy-1)+\lambda_2(y^2+z^2-9)[/tex]
with partial derivatives (set equal to 0)
[tex]L_x=y+\lambda_1y=y(1+\lambda_1)=0[/tex]
[tex]L_y=z+x+\lambda_1x+2\lambda_2y=0[/tex]
[tex]L_z=y+2\lambda_2z=0[/tex]
[tex]L_{\lambda_1}=xy-1=0[/tex]
[tex]L_{\lambda_2}=y^2+z^2-9=0[/tex]
[tex]xL_x=xy(1+\lambda_1)=1+\lambda_1=0\implies\lambda_1=-1[/tex]
[tex]\implies L_y=z+2\lambda_2y=0[/tex]
[tex]zL_y=z^2+2\lambda_2yz=0[/tex]
[tex]yL_z=y^2+2\lambda_2yz=0[/tex]
[tex]\implies zL_y+yL_z=9+4\lambda_2yz=0\implies\lambda_2yz=-\dfrac94[/tex]
[tex]\implies zL_y-yL_z=z^2-y^2=0\implies z^2=y^2\implies y=\pm z[/tex]
Now,
[tex]\lambda_2yz=-\dfrac94\impliesyL_z=y^2-\dfrac92=0\implies y=\pm\dfrac3{\sqrt2}\implies z=\mp\dfrac3{\sqrt2}[/tex]
[tex]y=\pm\dfrac3{\sqrt2}\implies x=\mp\dfrac{\sqrt2}3[/tex]
So we have four critical points to look into: [tex]\left(\dfrac2{\sqrt3},\dfrac3{\sqrt2},\dfrac3{\sqrt2}\right)[/tex], [tex]\left(-\dfrac2{\sqrt3},-\dfrac3{\sqrt2},-\dfrac3{\sqrt2}\right)[/tex], [tex]\left(\dfrac2{\sqrt3},\dfrac3{\sqrt2},-\dfrac3{\sqrt2}\right)[/tex], and [tex]\left(-\dfrac2{\sqrt3},-\dfrac3{\sqrt2},\dfrac3{\sqrt2}\right)[/tex].
Respectively, these give extreme values of [tex]\dfrac{11}2[/tex] (max), [tex]\dfrac{11}2[/tex] (max), [tex]-\dfrac72[/tex] (min), and [tex]-\dfrac72[/tex] (min).
Answer:
Extreme values [tex]\frac{11}{2} \rm max, \frac{11}{2} \rm max,\frac{-7}{2} \rm min,\frac{-7}{2} \rm min[/tex].
Step-by-step explanation:
Given information:
[tex]f(x,y,z)=yz+xy[/tex]
constraints condition:
[tex]\phi_1=xy-1=0[/tex]
[tex]\phi_2=y^2+z^2-9=0[/tex]
The LaGrange's auxiliary function
[tex]F=f(x,y,z)+ \lambda _1\phi_1+\lambda _2\phi_2[/tex]
[tex]F=yz+xy+\lambda_1(xy-1)+\lambda_2(y^2+z^2-9)=0[/tex]
On partial differentiation with respect to [tex]x[/tex] .
[tex]\frac{\partial F}{\partial x}=y+\lambda_1y=0=y(1+\lambda_1)=0[/tex] (equation number.1)
On partial differentiation with respect to [tex]y[/tex] .
[tex]\frac{\partial F}{\partial y}=z+x+\lambda_1x+2\lambda_2y=0[/tex] (equation number.2)
On partial differentiation with respect to [tex]z[/tex] .
[tex]\frac{\partial F}{\partial z}=y+2\lambda_2z=0[/tex] (equation number.3)
By (equation number 1)
[tex](1+\lambda_1)=0[/tex]
[tex]\lambda_1=-1[/tex]
By
[tex]z\frac{\partial F}{\partial y}=z^2+xz+\lambda_1xz+2\lambda_2yz=0[/tex]
[tex]y\frac{\partial F}{\partial z}=y^2+2\lambda_2yz=0[/tex]
On addition,
[tex]z\frac{\partial F}{\partial y}+y\frac{\partial F}{\partial z}=z^2+y^2+xz+\lambda_1xz+2\lambda_2yz+2\lambda_2yz=0[/tex]
put([tex]\lambda_1=-1[/tex]) and([tex]y^2+z^2=9[/tex])
[tex]\lambda_2yz=\frac{-9}{4}[/tex]
On subtraction,
[tex]z\frac{\partial F}{\partial y}-y\frac{\partial F}{\partial z}=z^2-y^2+xz+\lambda_1xz+2\lambda_2yz-2\lambda_2yz=0[/tex]
[tex]z^2-y^2=0\\z^2=y^2\\y=\pm z[/tex]
[tex]y+2\lambda_2z=0[/tex], [tex]\lambda_2yz=\frac{-9}{4}[/tex] from here,
[tex]y=\pm\frac{3}{\sqrt2}\\z=\mp\frac{3}{\sqrt2}[/tex]
[tex]\phi_1=xy-1=0[/tex],from here
[tex]x=\mp\frac{\sqrt2}{3}[/tex]
So we have four critical points,
[tex](\frac{2}{\sqrt3},\frac{3}{\sqrt2},\frac{3}{\sqrt2})(-\frac{2}{\sqrt3},-\frac{3}{\sqrt2},-\frac{3}{\sqrt2})(\frac{2}{\sqrt3},\frac{3}{\sqrt2},-\frac{3}{\sqrt2}) (-\frac{2}{\sqrt3},-\frac{3}{\sqrt2},\frac{3}{\sqrt2})[/tex]
Then put these value into the function
We get,
extreme values [tex]\frac{11}{2} \rm max, \frac{11}{2} \rm max,\frac{-7}{2} \rm min,\frac{-7}{2} \rm min[/tex].
Hence the answer is extreme values [tex]\frac{11}{2} \rm max, \frac{11}{2} \rm max,\frac{-7}{2} \rm min,\frac{-7}{2} \rm min[/tex].
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