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Albert1
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Given:
$x,y>0$ and $9x+4y=2005$, find:
$\min\left(\dfrac{1}{x}+\dfrac{1}{y}\right)$
$x,y>0$ and $9x+4y=2005$, find:
$\min\left(\dfrac{1}{x}+\dfrac{1}{y}\right)$
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Albert said:Given:
$x,y>0$ and $9x+4y=2005$, find:
$\min\left(\dfrac{1}{x}+\dfrac{1}{y}\right)$
Prove It said:$ \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 said:now what is the value of the minimum ?
The minimum value of $\dfrac{1}{x}+\dfrac{1}{y}$ is $\dfrac{1}{25}$.
To find the minimum value of $\dfrac{1}{x}+\dfrac{1}{y}$, we can use the method of Lagrange multipliers. This involves finding the critical points of the function $f(x,y)=\dfrac{1}{x}+\dfrac{1}{y}$ subject to the constraint $g(x,y)=9x+4y=2005$. The minimum value will be the smallest critical point that satisfies the constraint.
The constraints for finding the minimum value of $\dfrac{1}{x}+\dfrac{1}{y}$ are $x>0$, $y>0$, and $9x+4y=2005$. These constraints ensure that the values of $x$ and $y$ are positive and that they satisfy the given linear equation.
No, the minimum value of $\dfrac{1}{x}+\dfrac{1}{y}$ cannot be negative. Since $x$ and $y$ are both positive, the sum of their reciprocals will always be positive. The minimum value occurs when $x$ and $y$ are at their smallest possible values, which is $\dfrac{1}{25}$.
Yes, there are other methods for finding the minimum value of $\dfrac{1}{x}+\dfrac{1}{y}$, such as using calculus techniques like differentiation and optimization. However, the method of Lagrange multipliers is typically the most efficient and straightforward approach for solving this type of problem.