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zmudz
 one year ago
Let \(x\) and \(y\) be positive real numbers such that \(mx+y=1.\) Find the positive \(m\) such that the minimum of
\(\left( 1 + \frac{1}{x} \right)\left( 1 + \frac{m}{y} \right)\)
is 81.
zmudz
 one year ago
Let \(x\) and \(y\) be positive real numbers such that \(mx+y=1.\) Find the positive \(m\) such that the minimum of \(\left( 1 + \frac{1}{x} \right)\left( 1 + \frac{m}{y} \right)\) is 81.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0solve first for y and sub in the other equation to get only x. then rework equation to simplify. then you can differentiate, set equal to 81 and solve

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1can you use AMGM here because when you add the terms in brackets you get \(2 + \frac{1}{xy}\) so you end up solving something like \(\frac{1}{2}(2 + \frac{1}{xy})^2 > 81\) subject to \(y + mx = 1\) and funnily enough 81 is a square number

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1i know *no* pure maths.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1so could you say \(\frac{1}{\sqrt{2}}\sqrt{81} \lt \frac{1}{\sqrt{2}}\sqrt{\left( 1 + \frac{1}{x} \right)\left( 1 + \frac{m}{y} \right)}\lt \frac{1}{2} \left( 1 + \frac{1}{x} + 1 + \frac{m}{y} \right)\) and go from there?!?!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{1}{2}(81) \lt \frac{1}{4} \left( 2 + \frac{y+mx}{xy} \right)^2\] \[\left( 2 + \frac{1}{xy} \right) \gt 9 \sqrt{2} \]

zmudz
 one year ago
Best ResponseYou've already chosen the best response.1@freckles @Loser66 @ganeshie8 no idea how to differentiate  this is a from a precalc class. Is there a simpler way to solve this problem? Not sure what @IrishBoy123 is doing. Thanks!

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1\[ \left(1+\frac{1}{x}\right)\left(1+\frac{m}{y}\right)=1+\frac{1}{x}+\frac{m}{y}+\frac{m}{xy}\\ \frac{1}{4}\left(1+\frac{1}{x}+\frac{m}{y}+\frac{m}{xy}\right)\geq\sqrt[4]{\frac{m^2}{x^2y^2}}=\sqrt{\frac{m}{xy}}\\ \frac{1}{4}\left(1+\frac{1}{x}+\frac{m}{y}+\frac{m}{xy}\right)=\sqrt{\frac{m}{xy}}\quad \text{AMGM}\\ 1+\frac{1}{x}+\frac{m}{y}+\frac{m}{xy}=4\sqrt{\frac{m}{xy}}=81\\ 4\sqrt{\frac{m}{xy}}=81\\\\ \frac{m}{xy}=\frac{6561}{16}\\ mx+y=1\implies m<0\\ \frac{m}{xy}=\frac{6561}{16}=\frac{1}{x}=\frac{m}{y}=1!?\\ \] Furthermore, Mathematica shows that there is no unique m that satisfies the requirement. For every different x and y there is a different but unique m that satisfies it.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\begin{align} &\left(1+\frac{1}{x}\right)\left(1+\frac{m}{y}\right)\\ &=1+\frac{1}{x}+\frac{m}{y}+\frac{m}{xy}\\ &=1+\frac{y+mx+m}{xy}\\ &=1+\frac{1+m}{x(1mx)}\\ &\ge1+\frac{1+m}{1/(4m)}~~~\color{gray}{\because x(1mx)~\le~ 1/(4m) }\\ &=1+4m+4m^2\\ &=(2m+1)^2 \end{align}\] That means the minimum value of \(\left(1+\frac{1}{x}\right)\left(1+\frac{m}{y}\right)\) is \((2m+1)^2\) So, \((2m+1)^2= 81 \implies m =4,5 \). Since they want positive value, discard \(5\).

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1But \(m=4\) would certainly be incorrect since \(mx+y=1\) and \(x,y>0\).

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1Why \(\dfrac{y+mx+m}{xy}=\dfrac{1+m}{x(1mx)}\) and why \(x(1mx)\leq 4m\)?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I did my a super super long way... I differentiated: \[f(x)=(1+\frac{1}{x})(1+\frac{m}{1mx}) \\ \text{ and then set } f'(x) =0 \\ \text{ which eventually led me to the solution } x=\frac{1}{2m} \\ \text{ then I found } m \text{ such that } f(\frac{1}{2m})=81 \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2for first question, just replace \(y+mx\) by \(1\) for second question, since \(m\) is positive, \(x(1mx)\) is a parabola facing down with vertex at \((\frac{1}{2m}, \frac{1}{4m})\)
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