The demand functions for two products are $$p=12-2x\qquad \hbox{and}\qquad q=20-y$$

where $p$ and $q$ are the respective prices (in thousands of dollars) of each product, and $x$ and $y$ are the respective amounts (in thousands of units) of each sold. Suppose the joint cost function is $$C( x,y) =x^{2}+2xy+2y^{2}$$

Solution  (a) Revenue is the amount of money brought in. That is, revenue is the product of price and quantity sold. The revenue function $R$ is the sum of the revenues from each product. $$R( x,y) =xp+yq=x( 12-2x) +y( 20-y)$$

Profit is the difference between revenue and cost. The profit function $P$ is $$\begin{eqnarray*} P( x,y) &=&R( x,y) -C( x,y) =[ x(12-2x) +y( 20-y) ] -[ x^{2}+2xy+2y^{2}] \\[3pt] &=&12x-2x^{2}+20y-y^{2}-x^{2}-2xy-2y^{2}\\[3pt] &=&-3x^{2}-3y^{2}-2xy+12x+20y \end{eqnarray*}$$

(b) The partial derivatives of $P$ are $$\begin{eqnarray*} P_{x}( x,y) &=&\dfrac{\partial }{\partial x}(-3x^{2}-3y^{2}-2xy+12x+20y) =-6x-2y+12 \\[4pt] P_{y}( x,y) &=&\dfrac{\partial }{\partial y}(-3x^{2}-3y^{2}-2xy+12x+20y) =-6y-2x+20 \end{eqnarray*}$$

We find the critical points by solving the system of equations: $$\left\{ \begin{array}{l} -6x-2y+12=0 \\[3pt] -2x-6y+20=0 \end{array} \right.$$

We solve the first equation for $y$ and substitute the result into the second equation. Since $y=6-3x,$ then $$\begin{eqnarray*} -2x-6( 6-3x) +20 &=&0 \\ x &=&1 \end{eqnarray*}$$

and $y=6-3(1) =3$, so $(1,3)$ is the only critical point.

The second-order partial derivatives are $$A=P_{xx}( x,y) =-6\qquad B=P_{xy}( x,y) =-2\qquad C=P_{yy}( x,y) =-6$$

888

At the critical point $(1,3) ,$ $$\begin{array}{@{\hspace*{-3.4pc}}l} P_{xx}(1,3) =-6<0 \qquad P_{xy}(1,3) =-2 \qquad P_{yy}(1,3) =-6 \qquad AC-B^{2}=36-4=32>0 \end{array}$$

From the Second Partial Derivative Test, $P$ has a local maximum at $(1,3)$.

The domains of the demand functions $p=12-2x$ and $q=20-y$ are $0\leq x\leq 6$ and $0\leq y\leq 20$, respectively. Since the domain of the profit function $P$ is $0\leq x\leq 6,$ $0\leq y\leq 20,$ the domain is closed and bounded, so a maximum profit exists. From the test for absolute maximum and absolute minimum, the maximum profit is found at a critical point or on the boundary of the domain.

Figure 20 shows the boundary of the profit function $P$ and Table 1 gives the maximum value of $P$ on its boundary and at the critical point ($1,3$).

The boundary of $P$

We conclude that by selling $1000$ units of product $x$ for the price $p=12-2(1) =10$ thousand dollars and $3000$ units of product $y$ for the price $q=20-3=17$ thousand dollars, the profit is maximized.

(c) The maximum profit is $P(1,3) =$ 36,000.$