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For functions f(x, y, z) = xyz and g(x,y,z)=x2+2y2+3z21g(x, y, z)=x^{2}+2 y^{2}+3 z^{2}-1, write the Lagrange multiplier conditions that must be satisfied by a point that maximizes or minimizes f subject to the constraint g(x, y) = 0.


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Answered 2 years ago
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If we want to find a point P(x,y,z)P(x,y,z) that maximizes or minimizes the function f(x,y,z)f(x,y,z) subject to the constraint g(x,y,z)=0g(x,y,z)=0 we have to find x, y, zx,\ y,\ z and λ\lambda that satisfy the following Lagrange multipliers conditions:

f(x,y,z)=λg(x,y,z)andg(x,y,z)=0.\begin{align*} \nabla f(x,y,z)=\lambda \nabla g(x,y,z)\quad\text{and}\quad g(x,y,z)=0. \end{align*}

So first we need to find the gradients of the functions f(x,y,z)f(x,y,z) and g(x,y,z).g(x,y,z).

fx(x,y,z)=x(xyz)=yz;fy(x,y,z)=y(xyz)=xz;fz(x,y,z)=z(xyz)=xy;gx(x,y,z)=x(x2+2y2+3z21)=2x;gy(x,y,z)=y(x2+2y2+3z21)=4y;gz(x,y,z)=z(x2+2y2+3z21)=6z;f(x,y,z)=<fx,fy,fz>=<yz,xz,xy>g(x,y,z)=<gx,gy,gz>=<2x,4y,6z>.\begin{align*} f_x(x,y,z)&=\frac{\partial}{\partial x}(xyz)=yz;\\ f_y(x,y,z)&=\frac{\partial}{\partial y}(xyz)=xz;\\ f_z(x,y,z)&=\frac{\partial}{\partial z}(xyz)=xy;\\ g_x(x,y,z)&=\frac{\partial}{\partial x}(x^2+2y^2+3z^2-1)=2x;\\ g_y(x,y,z)&=\frac{\partial}{\partial y}(x^2+2y^2+3z^2-1)=4y;\\ g_z(x,y,z)&=\frac{\partial}{\partial z}(x^2+2y^2+3z^2-1)=6z;\\ \Longrightarrow\quad \nabla f(x,y,z)&=\left<f_x,f_y,f_z\right>=\left<yz,xz,xy\right>\\ \nabla g(x,y,z)&=\left<g_x,g_y,g_z\right>=\left<2x,4y,6z\right>. \end{align*}

The equation f(x,y,z)=λg(x,y,z)\nabla f(x,y,z)=\lambda \nabla g(x,y,z) can be rewritten as:

fx(x,y,z)=λgx(x,y,z)fy(x,y,z)=λgy(x,y,z)fz(x,y,z)=λgz(x,y,z)\begin{align*} f_x(x,y,z)&=\lambda g_x(x,y,z)\\ f_y(x,y,z)&=\lambda g_y(x,y,z)\\ f_z(x,y,z)&=\lambda g_z(x,y,z) \end{align*}

So the required Lagrange multipliers conditions are

yz=2λxxz=4λyxy=6λzx2+2y2+3z21=0.\begin{align*} yz=2\lambda x\\ xz=4\lambda y\\ xy=6\lambda z\\ x^2+2y^2+3z^2-1=0. \end{align*}

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