How do you minimize and maximize $f (x, y) = y e^{x} - x e^{y}$ $f(x,y)=ye^x-xe^y$ constrained to $x y = 4$ $xy=4$ ?

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How do you minimize and maximize $f (x, y) = y e^{x} - x e^{y}$ $f(x,y)=ye^x-xe^y$ constrained to $x y = 4$ $xy=4$ ?

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Answer 1 · 2017-12-30T17:07:07

$?$ $?$

Explanation:

$We could use the Lagrange multiplier L :$ $"We could use the Lagrange multiplier L :"$

$f (x, y, L) = y exp (x) - x exp (y) + L (x y - 4)$ $f(x,y,L) = y exp(x) - x exp(y) + L(xy - 4)$

$\frac{d f}{d x} = y exp (x) - exp (y) + L y = 0$ $(df)/(dx) = y exp(x) - exp(y) + L y = 0$
$\frac{d f}{d y} = exp (x) - x exp (y) + L x = 0$ $(df)/(dy) = exp(x) - x exp(y) + L x = 0$
$\frac{d f}{d L} = x y - 4 = 0 \Rightarrow y = \frac{4}{x}$ $(df)/(dL) = xy - 4 = 0 => y = 4/x$

$\Rightarrow (\frac{4}{x}) exp (x) - exp (\frac{4}{x}) + 4 \frac{L}{x} = 0$ $=> (4/x) exp(x) - exp(4/x) + 4 L / x = 0$
$\Rightarrow exp (x) - x exp (\frac{4}{x}) + L x = 0$ $=> exp(x) - x exp(4/x) + L x = 0$

$Multiply the last equation by (4/x) :$ $"Multiply the last equation by (4/x) : "$

$\Rightarrow (\frac{4}{x}) exp (x) - 4 exp (\frac{4}{x}) + 4 L = 0$ $=> (4/x) exp(x) - 4 exp(4/x) + 4 L = 0$

$Subtract this equation from the first :$ $"Subtract this equation from the first :"$

$\Rightarrow 3 exp (\frac{4}{x}) + 4 \frac{L}{x} - 4 L = 0$ $=> 3 exp(4/x) + 4 L / x - 4 L = 0$
$\Rightarrow 4 L (1 - \frac{1}{x}) = 3 exp (\frac{4}{x})$ $=> 4 L (1 - 1/x) = 3 exp(4/x)$
$\Rightarrow L = (\frac{3}{4}) \frac{exp (\frac{4}{x})}{1 - \frac{1}{x}}$ $=> L = (3/4) exp(4/x) / (1 - 1/x)$

$Fill in this value for L in the second equation :$ $"Fill in this value for L in the second equation :"$

$\Rightarrow exp (x) - x exp (\frac{4}{x}) + (\frac{3}{4}) exp (\frac{4}{x}) \frac{x^{2}}{x - 1} = 0$ $=> exp(x) - x exp(4/x) + (3/4) exp(4/x) x^2 / (x - 1) = 0$
$\Rightarrow exp (x) + exp (\frac{4}{x}) [(\frac{3}{4}) \frac{x^{2}}{x - 1} - x] = 0$ $=> exp(x) + exp(4/x) [ (3/4) x^2 / (x-1) - x ] = 0$
$\Rightarrow exp (x) = exp (\frac{4}{x}) \frac{x (x - 1) - (\frac{3}{4}) x^{2}}{x - 1}$ $=> exp(x) = exp(4/x) [ x(x - 1) - (3/4) x^2 ] / (x - 1)$
$\Rightarrow exp (x) = exp (\frac{4}{x}) x \frac{(\frac{1}{4}) x - 1}{x - 1}$ $=> exp(x) = exp(4/x) x [ (1/4) x - 1 ] / (x - 1)$
$\Rightarrow exp (x - \frac{4}{x}) = x \frac{(\frac{1}{4}) x - 1}{x - 1}$ $=> exp(x-4/x) = x [(1/4) x - 1]/(x - 1)$

$This looks like a transcendental equation.$ $"This looks like a transcendental equation."$
$I am stopping here, but if one solves this numerically, one gets$ $"I am stopping here, but if one solves this numerically, one gets"$
$x and then L and also y.$ $"x and then L and also y."$

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How do you minimize and maximize $f (x, y) = y e^{x} - x e^{y}$ $f(x,y)=ye^x-xe^y$ constrained to $x y = 4$ $xy=4$ ?

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