Finding the Min of $\dfrac{1}{x}+\dfrac{1}{y}$ Given $x,y>0$ and $9x+4y=2005$

[Albert1]

Given:

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

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

 

[Prove It]

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 $$\displaystyle \begin{align*} \frac{1}{x} + \frac{1}{y} \end{align*}$$ gives

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

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 :)

 

[Albert1]

$ \displaystyle \begin{align*} 9x + 4y &= 2005 \\ 4y &= -9x + 2005 \\ y &= \frac{-9x + 2005}{4} \end{align*} $

Substituting into the function $$\displaystyle \begin{align*} \frac{1}{x} + \frac{1}{y} \end{align*}$$ gives

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

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]

now what is the value of the minimum ?

You're not going to bring anything to the party? :P

 

[Albert1]

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 $\approx\dfrac{26}{2005}\approx0.013$

 

[Albert1]

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

 

[Albert1]

$ \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}$