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	{\noindent 21-120 Differential and Integral Calculus \hfill Summer Session I 2010\\}
	{\centering Quiz 9\medskip\\}
	\hrule
	{
		\begin{prob*}
			Find two positive intergers where the sum of the first and four times the second is 1000 and the product is as large as possible.
		\end{prob*}
		\begin{soln*}
			The algebra is
			\begin{gather*}
				P = xy \\
				x + 4y = 1000
			\end{gather*}
			Thus, using the second equation, we can eliminate a variable from the first. Thus we are left with maximizing
			\[ P = (1000 - 4y) y = 1000 y - 4y^2 \]
			Taking the first derivative, we have
			\[ P' = 1000 - 8y \]
			Setting this equal to 0, we see that $y = \frac{1000}{8} = 125$. The $x$ corresponding to this $y$ is $x=500$. We now only need to show this is a local max. To see that, we can observe that the parabola $1000y-4y^2$ is opening down, which means it must be a max. A more systematic way, however, is to take the second derivative
			\[ P'' = -8 \]
			Thus, this is concave down (looks like a frown) and must be a maximum.
		\end{soln*}
	}
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