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find minimum

Albert

Well-known member
Jan 25, 2013
1,225
Given:

$x,y>0$ and $9x+4y=2005$, find:

$\min\left(\dfrac{1}{x}+\dfrac{1}{y}\right)$
 
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Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Given:

$x,y>0$ and $9x+4y=2005$, find:

$\min\left(\dfrac{1}{x}+\dfrac{1}{y}\right)$
$ \displaystyle \begin{align*} 9x + 4y &= 2005 \\ 4y &= -9x + 2005 \\ y &= \frac{-9x + 2005}{4} \end{align*} $

Substituting into the function [tex]\displaystyle \begin{align*} \frac{1}{x} + \frac{1}{y} \end{align*} [/tex] gives

[tex]\displaystyle \begin{align*} f(x) &= \frac{1}{x} + \frac{1}{y} \\ &= \frac{1}{x} + \frac{1}{ \frac{-9x + 2005}{4} } \\ &= \frac{1}{x} + \frac{4}{-9x + 2005} \\ &= x^{-1} + 4 \left( -9x + 2005 \right)^{-1} \\ \\ f'(x) &= -x^{-2} + 36 \left( -9x + 2005 \right)^{-2} \\ &= -\frac{1}{x^2} + \frac{36}{\left( -9x + 2005 \right)^2 } \\ 0 &= -\frac{1}{x^2} + \frac{36}{\left( -9x + 2005 \right)^2} \textrm{ for a minimum } \\ \frac{1}{x^2} &= \frac{36}{\left( -9x + 2005 \right)^2} \\ \left( -9x + 2005 \right)^2 &= 36x^2 \\ 81x^2 - 36\, 090 + 4\, 020 \, 025 &= 36x^2 \\ 45x^2 - 36\, 090 + 4\, 020 \, 025 &= 0 \\ 9x^2 - 7218x + 804\,005 &= 0 \end{align*} [/tex]

Now solve this using the Quadratic Formula, and then substitute this value to find the value of y, and double-check that this is in fact a minimum :)
 

Albert

Well-known member
Jan 25, 2013
1,225
$ \displaystyle \begin{align*} 9x + 4y &= 2005 \\ 4y &= -9x + 2005 \\ y &= \frac{-9x + 2005}{4} \end{align*} $

Substituting into the function [tex]\displaystyle \begin{align*} \frac{1}{x} + \frac{1}{y} \end{align*} [/tex] gives

[tex]\displaystyle \begin{align*} f(x) &= \frac{1}{x} + \frac{1}{y} \\ &= \frac{1}{x} + \frac{1}{ \frac{-9x + 2005}{4} } \\ &= \frac{1}{x} + \frac{4}{-9x + 2005} \\ &= x^{-1} + 4 \left( -9x + 2005 \right)^{-1} \\ \\ f'(x) &= -x^{-2} + 36 \left( -9x + 2005 \right)^{-2} \\ &= -\frac{1}{x^2} + \frac{36}{\left( -9x + 2005 \right)^2 } \\ 0 &= -\frac{1}{x^2} + \frac{36}{\left( -9x + 2005 \right)^2} \textrm{ for a minimum } \\ \frac{1}{x^2} &= \frac{36}{\left( -9x + 2005 \right)^2} \\ \left( -9x + 2005 \right)^2 &= 36x^2 \\ 81x^2 - 36\, 090 + 4\, 020 \, 025 &= 36x^2 \\ 45x^2 - 36\, 090 + 4\, 020 \, 025 &= 0 \\ 9x^2 - 7218x + 804\,005 &= 0 \end{align*} [/tex]

Now solve this using the Quadratic Formula, and then substitute this value to find the value of y, and double-check that this is in fact a minimum :)
now what is the value of the minimum ?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403

Albert

Well-known member
Jan 25, 2013
1,225
The answer is $\dfrac{5}{401}=\dfrac{25}{2005}\approx0.0125$

Now I do it (in a quick way) only by estimation:

$\displaystyle \frac{1}{x}+\frac{1}{y}\ge2\sqrt{\frac{1}{xy}}= \frac{2}{x}$ it will happen when $x=y$.

$\therefore\,9x+4y=13x=2005\,\therefore\,x=\dfrac{2005}{13}$ and the minimum is $\approx\dfrac{26}{2005}\approx0.013$
 
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Albert

Well-known member
Jan 25, 2013
1,225
It is very strange the "LaTeX" in my post shows
normally at day time but reveals
"MATH EXPRESSION ERROR" at night
Can someone give me an aid ?
my best appreciation
Albert
 
Last edited:

Albert

Well-known member
Jan 25, 2013
1,225
$ \frac {1}{x}+\frac{1}{y}=\frac{2005+5y}{2005y-4y^2}=g(y)$

find the derivatives of g(y)=g'(y)

let g'(y)=0

we get :

$ 20y^2+8\times2005y-2005^2=0$

(10y-2005)(2y+2005)=0

$\therefore y=\frac {401}{2}\,\,,\,\, x= \frac {401}{3}\,\, (here\,\, x,y >0)$

$ min(\frac {1}{x}+\frac{1}{y})= =\frac {5}{401}$