Lagrange Multiplier mEthod

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Lagrange Multiplier mEthod

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man111

#1 h Jun 7, 2018, 9:19 PM • 2 Y

Y by Adventure10, Mango247

Finding maximum value of $x^p\cdot y^q$ . Given $x+y=1$ and $(x,y,p,q>0)$

Using Lagrange Multiplier Method.

This post has been edited 1 time. Last edited by man111, Jun 7, 2018, 9:20 PM

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#2 h Jun 7, 2018, 9:24 PM • 3 Y

Y by man111, Adventure10, Mango247

We must have $\nabla(x^py^q + \lambda(x + y - 1)) = \vec{0}$
And get ready to bash

This post has been edited 1 time. Last edited by whatRthose, Jun 7, 2018, 9:24 PM

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#3 h Apr 5, 2025, 11:15 AM

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Let us simply substitute $y=1-x$ to obtain $f(x) = x^{p}(1-x)^{q}$ for $x,y \in (0,1)$ and $p,q \in \mathbb{R}^{+}$ . Taking the derivative of $f$ equal to zero yields:

$f'(x) = px^{p-1}(1-x)^{q} - qx^{p}(1-x)^{q-1} = 0 \Rightarrow x^{p}(1-x)^{q}\left(\frac{p}{x}-\frac{q}{1-x}\right)=0 \Rightarrow x = 0, 1, \frac{p}{p+q}$ ;

of which only $x = \frac{p}{p+q}$ is admissible. A second derivative check of $f$ against this critical value produces:

$f''(x) = x^{p}(1-x)^{q}\left[\frac{p(p-1)}{x^2} - \frac{2pq}{x(1-x)} + \frac{q(q-1)}{(1-x)^2}\right] \Rightarrow f''\left(\frac{p}{p+q}\right) = -\frac{p^{p-1}q^{q-1}}{(p+q)^{p+q-3}} < 0 \Rightarrow$ a maximum for all $p,q > 0$ .

Hence, $f_{MAX} = f\left(\frac{p}{p+q}\right) = \textcolor{red}{\frac{p^{p}q^{q}}{(p+q)^{p+q}}}$ .

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#4 h Apr 6, 2025, 4:40 AM

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