Respuesta :
Answer:
Therefore the absolute maximum value is [tex]f(\frac43,\frac43)=8[/tex]
the absolute minimum value is [tex]f(-\frac43,-\frac43)=-8[/tex]
Step-by-step explanation:
Method of Lagrange multipliers:
1. Solve the following system of equation
[tex]\bigtriangledown L(x,y,..,\lambda) =\bigtriangledown f(x,y,..)+\lambda \bigtriangledown g(x,y,...)=0[/tex]
2. Plug the all solution (x,y,..) from the first step into f(x,y,..) and identify the maximum and minimum values.
The constant [tex]\lambda[/tex] is known as the Lagrange Multiplier.
Given function is
f(x,y)= 3x+3y
subject to constrain [tex]9x^2-9xy+9y^2=16[/tex]
[tex]g(x)=9x^2-9xy+9y^2-16=0[/tex]
The partial derivatives are
[tex]\frac{\partial f}{\partial x}=3[/tex]
[tex]\frac{\partial f}{\partial y}=3[/tex]
[tex]\frac{\partial g}{\partial x}=18x-9y[/tex]
[tex]\frac{\partial g}{\partial y}=-9x+18y[/tex]
[tex]\therefore \frac{\partial L}{\partial x}=3+\lambda (18x-9y)=0[/tex]......(1)
[tex]\therefore \frac{\partial L}{\partial y}=3+\lambda (-9x+18y)=0[/tex]......(2)
[tex]\therefore \frac{\partial L}{\partial \lambda}=9x^2-9xy+9y^2-16=0[/tex].....(3)
Multiplying 2 with equation (2) and add with equation (1)
[tex]3+ 18\lambda x-9\lambda y +6-18\lambda x+36\lambda y=0[/tex]
[tex]\Rightarrow 9+27 \lambda y=0[/tex]
[tex]\Rightarrow y=-\frac{9}{27 \lambda}[/tex]
[tex]\Rightarrow y=-\frac{1}{3 \lambda}[/tex]
Putting the value of y in equation (1)
[tex]3+ 18\lambda x-9\lambda (-\frac{1}{3\lambda})=0[/tex]
[tex]\Rightarrow 3+18\lambda x+3=0[/tex]
[tex]\Rightarrow x=-\frac{6}{18\lambda}[/tex]
[tex]\Rightarrow x=-\frac{1}{3 \lambda}[/tex]
Putting the value of x and y in equation (3)
[tex]9(-\frac{1}{3\lambda})^2-9(-\frac{1}{3\lambda})(-\frac{1}{3\lambda})+9(-\frac{1}{3\lambda})^2-16=0[/tex]
[tex]\Rightarrow 9(\frac{1}{9\lambda^2})-9(\frac{1}{9\lambda^2})+9(\frac{1}{9\lambda^2})-16=0[/tex]
[tex]\Rightarrow 1-1+1-16\lambda^2=0[/tex]
[tex]\Rightarrow \lambda^2=\frac{1}{16}[/tex]
[tex]\Rightarrow \lambda=\pm\frac{1}{4}[/tex]
Therefore the value of x and y becomes
[tex]x=\pm\frac{1}{3(\frac{1}{4})}[/tex]
[tex]\Rightarrow x=\pm \frac{4}{3}[/tex]
and [tex]y=\pm \frac{4}{3}[/tex]
Therefore
[tex]f(\frac{4}{3},\frac43) =(3\times \frac43)+(3\times \frac43)[/tex]
=4+4
=8
[tex]f(-\frac{4}{3},-\frac43) =[3\times (-\frac43)]+[3\times(- \frac43)][/tex]
=-4-4
=-8
Therefore the absolute maximum value is [tex]f(\frac43,\frac43)=8[/tex]
the absolute minimum value is [tex]f(-\frac43,-\frac43)=-8[/tex]
