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\textbf{Applications of Partial Differentiation}

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

Find the three positive numbers $a$, $b$ and $c$ given that the sum of
these numbers is 30 and for which the expression $ab^2c^3$ is maximum.


\textbf{Answer}

It is given that
$$a&gt;0, \ b&gt;0, \ c&gt;0$$
$$\rm{and} \ \ \ a+b+c=30$$
and we want to maximize the following:
\begin{eqnarray*}
P &amp; = &amp; ab^2c^3 = (30-b-c)b^2c^3\\
&amp; = &amp; 30b^2c^3-b^3c^3-b^2c^4
\end{eqnarray*}
Since $P=0$ if $b=0$ or $c=0$ or $b+c=30$ (i.e. $a=30$), the maximum
value of $P$ will occur at a critical point $(b,c)$ satisfying $b&gt;0$,
$c&gt;0$ and $b+c&lt;30$.

For CP:
\begin{eqnarray*}
0 = \frac{\partial P}{\partial b} &amp; = &amp; 60bc^3-3b^2c^3-2bc^4\\
&amp; = &amp; bc^3(60-3b-2c)\\
0= \frac{\partial P}{\partial c} &amp; = &amp; 90b^2c^2-3b^3c^2-4b^2c^3\\
&amp; = &amp; b^2c^2(90-3b-4c)
\end{eqnarray*}
Hence $9b+6c=180=6b+8c$, from which we obtain $3b=2c=30$.

The three numbers are $b=10$, $c=15$ and $a=30-10-15=5$.


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