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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=xy127x2y3xP=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?


Answered 5 months ago
Answered 5 months ago
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If xx thousand widgets are manufactured per month and yy number of widgets are sold. then profit per month

P=xyx2y327x\begin{align*} P & = xy - \dfrac{x^2y^3}{27} - x \end{align*}

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

0x3, 0y2\begin{align*} 0 \le x \le 3\,, & \ 0 \le y \le 2 \end{align*}

To maximize PP, we have to find critical points. Therefore,

Px=0x[xyx2y327x]=0[y2xy3271]=0\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

Py=0y[xyx2y327x]=0[xx2y29]=0x(1xy29)=0      x=0,(1xy29)=0\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

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

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

Solving the above given simultaneous equations in xx and yy, we have

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

Hence, the critical point is (1,3)(1,3). But this point not inside the constraints region RR i.e. 0x3, 0y20\le x \le 3, \ 0 \le y \le 2.

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