Respuesta :
Answer:
[tex]\displaystyle x= -\frac{10}{\sqrt{269}}\\\\\displaystyle y= \frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{23\sqrt{269}+269}{269}[/tex]
Maximum value of f=2.41
Step-by-step explanation:
Lagrange Multipliers
It's a method to optimize (maximize or minimize) functions of more than one variable subject to equality restrictions.
Given a function of three variables f(x,y,z) and a restriction in the form of an equality g(x,y,z)=0, then we are interested in finding the values of x,y,z where both gradients are parallel, i.e.
[tex]\bigtriangledown f=\lambda \bigtriangledown g[/tex]
for some scalar [tex]\lambda[/tex] called the Lagrange multiplier.
For more than one restriction, say g(x,y,z)=0 and h(x,y,z)=0, the Lagrange condition is
[tex]\bigtriangledown f=\lambda \bigtriangledown g+\mu \bigtriangledown h[/tex]
The gradient of f is
[tex]\bigtriangledown f=<f_z,f_y,f_z>[/tex]
Considering each variable as independent we have three equations right from the Lagrange condition, plus one for each restriction, to form a 5x5 system of equations in [tex]x,y,z,\lambda,\mu[/tex].
We have
[tex]f(x, y, z) = x + 2y + 11z\\g(x, y, z) = x - y + z -1=0\\h(x, y, z) = x^2 + y^2 -1= 0[/tex]
Let's compute the partial derivatives
[tex]f_x=1\ ,f_y=2\ ,f_z=11\ \\g_x=1\ ,g_y=-1\ ,g_z=1\\h_x=2x\ ,h_y=2y\ ,h_z=0[/tex]
The Lagrange condition leads to
[tex]1=\lambda (1)+\mu (2x)\\2=\lambda (-1)+\mu (2y)\\11=\lambda (1)+\mu (0)[/tex]
Operating and simplifying
[tex]1=\lambda+2x\mu\\2=-\lambda +2y\mu \\\lambda=11[/tex]
Replacing the value of [tex]\lambda[/tex] in the two first equations, we get
[tex]1=11+2x\mu\\2=-11 +2y\mu[/tex]
From the first equation
[tex]\displaystyle 2\mu=\frac{-10}{x}[/tex]
Replacing into the second
[tex]\displaystyle 13=y\frac{-10}{x}[/tex]
Or, equivalently
[tex]13x=-10y[/tex]
Squaring
[tex]169x^2=100y^2[/tex]
To solve, we use the restriction h
[tex]x^2 + y^2 = 1[/tex]
Multiplying by 100
[tex]100x^2 + 100y^2 = 100[/tex]
Replacing the above condition
[tex]100x^2 + 169x^2 = 100[/tex]
Solving for x
[tex]\displaystyle x=\pm \frac{10}{\sqrt{269}}[/tex]
We compute the values of y by solving
[tex]13x=-10y[/tex]
[tex]\displaystyle y=-\frac{13x}{10}[/tex]
For
[tex]\displaystyle x= \frac{10}{\sqrt{269}}[/tex]
[tex]\displaystyle y= -\frac{13}{\sqrt{269}}[/tex]
And for
[tex]\displaystyle x= -\frac{10}{\sqrt{269}}[/tex]
[tex]\displaystyle y= \frac{13}{\sqrt{269}}[/tex]
Finally, we get z using the other restriction
[tex]x - y + z = 1[/tex]
Or:
[tex]z = 1-x+y[/tex]
The first solution yields to
[tex]\displaystyle z = 1-\frac{10}{\sqrt{269}}-\frac{13}{\sqrt{269}}[/tex]
[tex]\displaystyle z = \frac{-23\sqrt{269}+269}{269}[/tex]
And the second solution gives us
[tex]\displaystyle z = 1+\frac{10}{\sqrt{269}}+\frac{13}{\sqrt{269}}[/tex]
[tex]\displaystyle z = \frac{23\sqrt{269}+269}{269}[/tex]
Complete first solution:
[tex]\displaystyle x= \frac{10}{\sqrt{269}}\\\\\displaystyle y= -\frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{-23\sqrt{269}+269}{269}[/tex]
Replacing into f, we get
f(x,y,z)=-0.4
Complete second solution:
[tex]\displaystyle x= -\frac{10}{\sqrt{269}}\\\\\displaystyle y= \frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{23\sqrt{269}+269}{269}[/tex]
Replacing into f, we get
f(x,y,z)=2.4
The second solution maximizes f to 2.4
