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Question

# A widget manufacturer determines that if she manufactures x thousands of widgets per month and sells the widgets for y dollars each, then her monthly profit (in thousands of dollars) will be $P=x y-\frac{1}{27} x^{2} y^{3}-x$. If her factory is capable of producing at most 3,000 widgets per month, and government regulation s prevent her from charging more than \$2 per widget, how many should she manufacture, and how much should she charge for each, to maximize her monthly profit?

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If $x$ thousand widgets are manufactured per month and $y$ number of widgets are sold. then profit per month

\begin{align*} P & = xy - \dfrac{x^2y^3}{27} - x \end{align*}

We have to maximize the profit. The constraints are given as

\begin{align*} 0 \le x \le 3\,, & \ 0 \le y \le 2 \end{align*}

To maximize $P$, we have to find critical points. Therefore,

\begin{align*} \dfrac{\partial P}{\partial x} & = 0 \\ \dfrac{\partial }{\partial x}\left[xy - \dfrac{x^2y^3}{27} - x \right] & = 0 \\ \left[y - \dfrac{2x y^3}{27} - 1 \right] & = 0 \end{align*}

Similarly, we have

\begin{align*} \dfrac{\partial P}{\partial y} & = 0 \\ \dfrac{\partial }{\partial y}\left[xy - \dfrac{x^2y^3}{27} - x \right] & = 0 \\ \left[x - \dfrac{x^2y^2}{9}\right] & = 0 \\ x\left(1 - \dfrac{xy^2}{9}\right) & = 0 \ \implies \ x = 0, \left(1 - \dfrac{xy^2}{9}\right)=0 \end{align*}

Here, we see that

\begin{align*} x & = 0 \end{align*}

$x=0$ is not feasible as it gives zero profit.

Solving the above given simultaneous equations in $x$ and $y$, we have

\begin{align*} x & = 1 \\ y & = 3 \end{align*}

Hence, the critical point is $(1,3)$. But this point not inside the constraints region $R$ i.e. $0\le x \le 3, \ 0 \le y \le 2$.

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