Answer with Step-by-step explanation:
We are given that
[tex]f(x,y,z)=10x+10y+3z[/tex]
[tex]g(x,y,z)=5x^2+5y^2+3z^2=43[/tex]
We have to find the extreme values of the function.
[tex]f_x(x,y,z)=10,f_y(x,y,z)=10,f_z(x,y,z)=3[/tex]
[tex]g_x(x,y,z)=10x,g_y(x,y,z)=10y,g_z(x,y,z)=6z[/tex]
[tex]\Delta f(x,y,z)=\lambda \Delta g(x,y,z)[/tex]
Therefore,
[tex]10=\lambda 10x\implies x=\frac{1}{\lambda}[/tex]
[tex]10=10y\lambda\implies y=\frac{1}{\lambda}[/tex]
[tex]3=6z\lambda[/tex]
[tex]z=\frac{1}{2\lambda}[/tex]
Substitute the values in g(x)
Then, we get
[tex]\frac{5}{\lambda^2}+\frac{5}{\lambda^2}+\frac{3}{4\lambda^2}=43[/tex]
[tex]\frac{20+20+3}{4\lambda^2}=43[/tex]
[tex]\frac{43}{4\lambda^2}=43[/tex]
[tex]\lambda^2=\frac{43}{4\times 43}[/tex]
[tex]\lambda=\pm\frac{1}{2}[/tex]
When [tex]\lambda=\frac{1}{2}[/tex] then,
[tex]x=2,y=2,z=1[/tex]
When [tex]\lambda=-\frac{1}{2}[/tex]
Then, [tex]x=-2,y=-2,z=-1[/tex]
[tex]f(2,2,1)=10+10+3=23[/tex]
[tex]f(-2,-2,-1)=-10-10-3=-23[/tex]
Hence, maximum value of f(x,y,z) is 23 at (2,2,1) and minimum value of f(x,y,z) is -23 at (-2,-2 ,-1).
