## 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.

1. anonymous

solve 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

2. IrishBoy123

can you use AM-GM 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

3. IrishBoy123

i know *no* pure maths.

4. IrishBoy123

so 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?!?!

5. IrishBoy123

$\frac{1}{2}(81) \lt \frac{1}{4} \left( 2 + \frac{y+mx}{xy} \right)^2$ $\left( 2 + \frac{1}{xy} \right) \gt 9 \sqrt{2}$

6. zmudz

@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!

7. thomas5267

$\left(1+\frac{1}{x}\right)\left(1+\frac{m}{y}\right)=1+\frac{1}{x}+\frac{m}{y}+\frac{m}{xy}\\ \frac{1}{4}\left(1+\frac{1}{x}+\frac{m}{y}+\frac{m}{xy}\right)\geq\sqrt[4]{\frac{m^2}{x^2y^2}}=\sqrt{\frac{|m|}{xy}}\\ \frac{1}{4}\left(1+\frac{1}{x}+\frac{m}{y}+\frac{m}{xy}\right)=\sqrt{\frac{|m|}{xy}}\quad \text{AM-GM}\\ 1+\frac{1}{x}+\frac{m}{y}+\frac{m}{xy}=4\sqrt{\frac{|m|}{xy}}=81\\ 4\sqrt{\frac{|m|}{xy}}=81\\\\ \frac{|m|}{xy}=\frac{6561}{16}\\ mx+y=1\implies m<0\\ \frac{m}{xy}=\frac{-6561}{16}=\frac{1}{x}=\frac{m}{y}=1!?\\$ 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.

8. ganeshie8

\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(1-mx)}\\ &\ge1+\frac{1+m}{1/(4m)}~~~\color{gray}{\because x(1-mx)~\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 $$-5$$.

9. thomas5267

But $$m=4$$ would certainly be incorrect since $$mx+y=1$$ and $$x,y>0$$.

10. ganeshie8

try x = 1/8 y = 1/2

11. ganeshie8
12. thomas5267

Oh I see!

13. thomas5267

Why $$\dfrac{y+mx+m}{xy}=\dfrac{1+m}{x(1-mx)}$$ and why $$x(1-mx)\leq 4m$$?

14. freckles

I did my a super super long way... I differentiated: $f(x)=(1+\frac{1}{x})(1+\frac{m}{1-mx}) \\ \text{ and then set } f'(x) =0 \\ \text{ which eventually led me to the solution } x=\frac{1}{2m} \\ \text{ then I found } m \text{ such that } f(\frac{1}{2m})=81$

15. ganeshie8

for first question, just replace $$y+mx$$ by $$1$$ for second question, since $$m$$ is positive, $$x(1-mx)$$ is a parabola facing down with vertex at $$(\frac{1}{2m}, \frac{1}{4m})$$