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\textbf{Applications of Partial Differentiation}

\textit{\textbf{Extremes}}
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\textbf{Question}

Find the minimum value of $$f(x,y) = x+8y +\frac{1}{xy}$$ in the first
quadrant ($x>0$, $y>0$).

How do you know that a minimum exists?


\textbf{Answer}

$\begin{array}{rlrl}
f(x,y) & =  \displaystyle x+8y +\frac{1}{xy} & & x>0, \ y>0\\
f_1(x,y) & =  \displaystyle 1 - \frac{1}{x^2y} =0 & \Rightarrow &
x^2y=1\\
f_2(x,y) & =  \displaystyle 8 - \frac{1}{xy^2} =0 7 \Rightarrow &
8xy^2=1
\end{array}$

The critical points must satisfy $$\frac{x}{y} = \frac{x^2y}{xy^2}=8$$
that is $x=8y$. Also, $x^2y = 1$, so $64y^3=1$.

Thus $y=1/4$ and $x=2$; the critical point is $(2,\frac{1}{4})$.

Since $f(x,y)\to\infty$ if $x\to 0+$, $y\to 0+$, or
$x^2+y^2\to\infty$, the critical point must give a minimum value for
$f$.

The minimum value is $f(2,\frac{1}{4}) =2+2+2=6$.

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