Answer by Doug M for Find extreme values of $f$ subject to both constraints....
$x = y\\2x+z = 24\\2x^2 = z^2\\x = \pm \frac {\sqrt 2}{2} z$$(1+\frac {\sqrt 2}{2}) z = 24, (1-\frac {\sqrt 2}{2}) z = 24\\z = \frac {48}{2+\sqrt2}, z = \frac {48}{2-\sqrt2}\\z = \frac 24(2-\sqrt2), z...
View ArticleFind extreme values of $f$ subject to both constraints. $f=z; g=x^2+y^2=z^2;...
$f=z \\g=x^2+y^2=z^2\\ h=x+y+z=24$I'm having a hard time soloing out $\mu, \lambda, x, y,$ or $z$.$\triangle(f)=\langle0,0,1\rangle\\\triangle(g)=\langle2x, 2y,...
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