Respuesta :
The expression
10 x + 10 y + 4 z + λ (5 x^2 + 5 y^2 + 4 z^2 - 44)
has partial derivatives with respect to x, y, z, and λ of
10 + 10 x λ
10 + 10 y λ
4 + 8 z λ
-44 + 5 x^2 + 5 y^2 + 4 z^2
Setting these to zero and solving simultaneously gives the solutions
(x, y, z, λ) = (2, 2, 1, -1/2)
(x, y, z, λ) = (-2, -2, -1, 1/2)
The corresponding extreme values of f(x, y, z) are 44 and -44.
10 x + 10 y + 4 z + λ (5 x^2 + 5 y^2 + 4 z^2 - 44)
has partial derivatives with respect to x, y, z, and λ of
10 + 10 x λ
10 + 10 y λ
4 + 8 z λ
-44 + 5 x^2 + 5 y^2 + 4 z^2
Setting these to zero and solving simultaneously gives the solutions
(x, y, z, λ) = (2, 2, 1, -1/2)
(x, y, z, λ) = (-2, -2, -1, 1/2)
The corresponding extreme values of f(x, y, z) are 44 and -44.
The extreme value of the given function is [tex]\boxed{\bf 44\text{ and }-44}[/tex].
Further explanation:
Given:
The function to the constraint is,
[tex]\boxed{f(x)=10x+10y+4z+\lambda(5x^{2}+5y^{2}+4z^{2}-44)}[/tex]
Calculation:
The given function is as follows:
[tex]\boxed{f(x)=10x+10y+4z+\lambda(5x^{2}+5y^{2}+4z^{2}-44)}[/tex]
Differentiate the function [tex]f(x)[/tex] with respect to [tex]x[/tex] as,
[tex]\boxed{\dfrac{\partial {f(x)}}{\partial x}=10+10x\lambda}[/tex] ....(1)
Substitute [tex]\frac{\partial f(x)}{\partial x}=0[/tex] in the equation (1) as follows:
[tex]\begin{aligned}10+10x\lambda&=0\\x\lambda&=-1\\x&=-\dfrac{1}{\lambda}\\x^{2}&=\dfrac{1}{\lambda^{2}}\end{aligned}[/tex]
Differentiate the function [tex]f(x)[/tex] with respect to [tex]y[/tex] as follows:
[tex]\boxed{\dfrac{\partial f(x)}{\partial y}=10+10y\lambda}[/tex] .....(2)
Substitute [tex]\frac{\partial f(x)}{\partial y}=0[/tex] in the equation (2) as follows:
[tex]\begin{aligned}10+10y\lambda&=0\\y\lambda&=-1\\y^{2}&=\dfrac{1}{\lambda^{2}}\end{aligned}[/tex]
Differentiate the function [tex]f(x)[/tex] with respect to [tex]z[/tex] as follows:
[tex]\boxed{\dfrac{\partial f(x)}{\partial z}=10+10z\lambda}[/tex] .....(3)
Substitute [tex]\frac{\partial f(x)}{\partial z}=0[/tex] in the equation (3) as follows:
[tex]\begin{aligned}4+8z\lamnda&=0\\8z\lambda&=-4\\z\lambda&=\dfrac{-4}{8}\\z\lambda&=-\dfrac{1}{2}\\z^{2}&=\dfrac{1}{2\lambda^{2}}\end{aligned}[/tex]
Differentiate the function [tex]f(x)[/tex] with respect to [tex]\lambda[/tex] as follows:
[tex]\boxed{\dfrac{\partial f(x)}{\partial \lambda}=5x^{2}+5y^{2}+4z^{2}-44}[/tex] .....(4)
Substitute [tex]\frac{\partial f(x)}{\partial \lambda}=0[/tex] in the equation (4) as follows:
[tex]\begin{aligned}5x^{2}+5y^{2}+4z^{2}-44&=0\\5x^{2}+5y^{2}+4z^{2}&=44\end{aligned}[/tex]
Substitute [tex]x^{2}=\frac{1}{\lambda^{2}}[/tex], [tex]y^{2}=\frac{1}{\lambda^{2}}[/tex] and [tex]z^{2}=\frac{1}{\lambda^{2}}[/tex] in the above equation as follows:
[tex]\begin{aligned}\dfrac{5}{\lambda^{2}}+\dfrac{5}{\lambda^{2}}+\dfrac{4}{4\lambda^{2}}&=44\\ \dfrac{5}{\lambda^{2}}+\dfrac{5}{\lambda^{2}}+\dfrac{1}{\lambda^{2}}&=44\\ \dfrac{11}{\lambda^{2}}&=44\\ \lambda^{2}&=\dfrac{11}{44}\end{aligned}[/tex]
Simplify the above equation as follows:
[tex]\begin{aligned}\lambda^{2}&=\dfrac{1}{4}\\ \lamda&=\pm \sqrt{\dfrac{1}{4}}\\&=\pm \dfrac{1}{2}\end{aligned}[/tex]
The value of [tex]\lambda[/tex] is [tex]\pm \frac{1}{2}[/tex].
Substitute [tex]\lambda=\frac{1}{2}[/tex] in the equation [tex]x\lambda=-1[/tex] as follows:
[tex]\begin{aligned}x\cdot \dfrac{1}{2}&=-1\\x&=-2\end{aligned}[/tex]
Now, substitute [tex]\lambda=-\frac{1}{2}[/tex] in the equation [tex]x\lambda=-1[/tex] as follows:
[tex]\begin{aligned}x\cdot \dfrac{-1}{2}&=-1\\x&=2\end{aligned}[/tex]
Substitute [tex]\lambda=\frac{1}{2}[/tex] in the equation [tex]y\lambda=-1[/tex] as follows:
[tex]\begin{aligned}y\cdot \dfrac{1}{2}&=-1\\y&=-2\end{aligned}[/tex]
Substitute [tex]\lambda=-\frac{1}{2}[/tex] in the equation [tex]y\lambda=-1[/tex] as follows:
[tex]\begin{aligned}y\cdot \dfrac{-1}{2}&=-1\\y&=2\end{aligned}[/tex]
Substitute [tex]\lambda=\frac{1}{2}[/tex] in the equation [tex]z\lambda=\frac{-1}{2}[/tex] as follows:
[tex]\begin{aligned}z\cdot \dfrac{1}{2}&=-\dfrac{1}{2}\\z&=-1\end{aligned}[/tex]
Now, substitute [tex]\lambda=\frac{-1}{2}[/tex] in the equation [tex]z\lambda=\frac{-1}{2}[/tex] as follows:
[tex]\begin{aligned}z\cdot \dfrac{-1}{2}&=-\dfrac{1}{2}\\z&=1\end{aligned}[/tex]
Therefore, the values of [tex](x,y,x,\lambda)[/tex] are [tex](2,2,1,\frac{-1}{2})[/tex] and [tex](-2,-2,-1,\frac{1}{2})[/tex].
Substitute [tex]2[/tex] for [tex]x[/tex],[tex]2[/tex] for [tex]y[/tex],[tex]1[/tex] for [tex]z[/tex] and [tex]\frac{-1}{2}[/tex] for [tex]\lambda[/tex] in the function [tex]f(x)[/tex] as follows,
[tex]\begin{aligned}f(x)&=(10\cdot 2)+(10\cdot 2)+(4\cdot 1)-\dfrac{1}{2}((5\cdot 4)+(5\cdot 4)+(4\cdot 1)-44)\\&=44-\dfrac{1}{2}\cdot (0)\\&=44\end{aligned}[/tex]
Therefore, the maximum value of the function is [tex]\boxed{\bf 44}[/tex].
Substitute [tex]-2[/tex] for [tex]x[/tex],[tex]-2[/tex] for [tex]y[/tex],[tex]-1[/tex] for [tex]z[/tex] and [tex]\frac{1}{2}[/tex] for [tex]\lambda[/tex] in the function [tex]f(x)[/tex] as follows:
[tex]\begin{aligned}f(x)&=(10\cdot (-2))+(10\cdot (-2))+(4\cdot (-1))-\dfrac{1}{2}((5\cdot 4)+(5\cdot 4)+(4\cdot 1)-44)\\&=-44-\dfrac{1}{2}\cdot (0)\\&=-44\end{aligned}[/tex]
Therefore, the minimum value of the function is [tex]\boxed{\bf -44}[/tex].
Learn more:
1. Learn more about problem on numbers: https://brainly.com/question/1852063
2. Learn more about problem on function https://brainly.com/question/3225044
Answer details:
Grade: Senior school
Subject: Mathematics
Chapter: Differentiation
Keywords: Extreme values, maximum value, minimum value, function, constraint, differentiate, 44, -44, lagrange multipliers, solution, f(x,y,z)=10x+10y+4z, 5x2+5y2+4z2=44.
