Using Lagrange multipliers, show that: a) maximum value of $x^2y^3z^4$ subject to $2x+3y+4z=a$ is $(a/9)^9$

Question

Using Lagrange multipliers, show that:
a) maximum value of $x^2y^3z^4$ subject to $2x+3y+4z=a$ is $(a/9)^9$
b) minimum value of $yz+zx+xy$ subject to $xyz=a^2(x+y+z)$ is $9a^2$

Solution

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a) maximum value of $x^2y^3z^4$ subject to $2x+3y+4z=a$ is $(a/9)^9$

Objective function, $f=x^2y^3z^4$
Constraint function, $g=2x+3y+4z-a$
Lagrange multiplier function: $\nabla f(x_0,y_0,z_0)=\lambda_0 \nabla g(x_0,y_0,z_0)$

Step-1: Find Gradient of f

$\nabla f(x_0,y_0,z_0) = \begin{bmatrix} \frac{\partial f}{\partial x}\\ \frac{\partial f}{\partial y}\\ \frac{\partial f}{\partial z}\end{bmatrix} = \begin{bmatrix} 2xy^3z^4\\ 3y^2x^2z^4\\ 4z^3x^2y^3 \end{bmatrix}$

Step-2: Find Gradient of g

$\nabla g(x_0,y_0,z_0) = \begin{bmatrix} \frac{\partial g}{\partial x}\\ \frac{\partial g}{\partial y}\\ \frac{\partial g}{\partial z}\end{bmatrix} = \begin{bmatrix} 2\\ 3\\ 4 \end{bmatrix}$

Step-3: Find x, y and $\lambda_0$

$\begin{bmatrix} 2xy^3z^4\\ 3y^2x^2z^4\\ 4z^3x^2y^3 \end{bmatrix} = \lambda_0 \begin{bmatrix} 2\\ 3\\ 4 \end{bmatrix}$

$\Rightarrow \lambda_0 = x^8 = y^8 = z^8$

$\Rightarrow x = y = z = \pm \lambda_0^{1/8}$

Step-4: Substitute and figure out function values
Let’s take $\lambda_0^{1/8}$ as the question is to prove maxima.

Let c = $\lambda_0^{1/8}$ ,

Apply c in Constraint function $\Rightarrow$

$g(x,y,z)=2x+3y+4z-a \Rightarrow 2c+3c+4c-a=9c-a=0 \Rightarrow \lambda_0^{1/8} = a/9$

Apply (7) in (1) $\Rightarrow$

$f(x,y,z)=x^2y^3z^4=(a/9)^2.(a/9)^3.(a/9)^4=(a/9)^9.$

Thus, f has maxima at $(a/9)^9$ . Hence, proved.
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b) minimum value of $yz+zx+xy$ subject to $xyz=a^2(x+y+z)$ is $9a^2$

Objective function, $f(x,y,z)=yz+zx+xy$
Constraint function, $g(x,y,z)=xyz−a^2(x+y+z)$
Lagrange multiplier function: $\nabla f(x_0,y_0,z_0)=\lambda_0 \nabla g(x_0,y_0,z_0)$

Step-1: Find Gradient of f

$\nabla f(x_0,y_0,z_0) = \begin{bmatrix} \frac{\partial f}{\partial x}\\ \frac{\partial f}{\partial y}\\ \frac{\partial f}{\partial z}\end{bmatrix} = \begin{bmatrix} y+z\\ x+z\\ x+y \end{bmatrix}$

Step-2: Find Gradient of g

$\nabla g(x_0,y_0,z_0) = \begin{bmatrix} \frac{\partial g}{\partial x}\\ \frac{\partial g}{\partial y}\\ \frac{\partial g}{\partial z}\end{bmatrix} = \begin{bmatrix} yz-a^2\\ xz-a^2\\ xy-a^2 \end{bmatrix}$

Step-3: Find x, y and $\lambda_0$

$\Rightarrow \begin{bmatrix} y+z\\ x+z\\ x+y \end{bmatrix} = \lambda_0 \begin{bmatrix} yz-a^2\\ xz-a^2\\ xy-a^2 \end{bmatrix}$

(1) $\begin{equation*} \left\ y+z = \lambda_0(yz-a^2) \Rightarrow y+z-\lambda_0yz = -\lambda_0 a^2 \end{equation*}$

Similarly, we get,

(2) $\begin{equation*} \left\ \Rightarrow x+z-\lambda_0xz = -\lambda_0 a^2 \end{equation*}$

(3) $\begin{equation*} \left\ \Rightarrow x+y-\lambda_0xy = -\lambda_0 a^2 \end{equation*}$

RHS of the equations (1), (2) and (3) are the same. Thus, let’s equate LHS of (1) = (2)

$y+z-\lambda_0yz=x+z-\lambda_0xz \Rightarrow x-y=\lambda_0z(x−y) \Rightarrow \lambda_0=1/y$

Similarly, (2)=(3) and (1)=(3) will give $\lambda_0$ =1/x, $\lambda_0$ =1/z
Thus, x=y=z= $1/\lambda_0$

Step-4: Substitute $\lambda_0$ and figure out function values
Apply $\lambda_0$ in Constraint function $\Rightarrow$
Let $c=1/\lambda_0$

$g(x,y,z)=xyz−a^2(x+y+z) \Rightarrow c^3=3c.a^2 \Rightarrow c^2=3a^2 \Rightarrow c = \pm \sqrt{3}a \Rightarrow 1/\lambda_0 = \pm \sqrt{3}a$

Let’s take $-\sqrt{3}a$ as the question is about a minima.

Apply $\lambda_0$ in Objective function $\Rightarrow$

$f(x,y,z) = yz+zx+xy = c^2+c^2+c^2=3c^2=3/\lambda_0^2 \Rightarrow f(x,y,z)=9a^2$

Thus, f has minima at $9a^2$ . Hence, proved.

Question

Solution

a) maximum value of $x^2y^3z^4$ subject to $2x+3y+4z=a$ is $(a/9)^9$

b) minimum value of $yz+zx+xy$ subject to $xyz=a^2(x+y+z)$ is $9a^2$

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Question

Solution

a) maximum value of subject to is

b) minimum value of subject to is

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a) maximum value of $x^2y^3z^4$ subject to $2x+3y+4z=a$ is $(a/9)^9$

b) minimum value of $yz+zx+xy$ subject to $xyz=a^2(x+y+z)$ is $9a^2$