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 \[ A=P_{xx}( x,y) =-6\qquad B=P_{xy}( x,y) =-2\qquad C=P_{yy}( x,y) =-6 \]
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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 (\(1,3\)).
| Maximum \(\bf P\) | ||
|---|---|---|
| \(L_{1}:\quad 0\le x \le 6, y=0\) | \(P(x,0)= -3x^{2}+12x\) | \(P(2,0)=12\) |
| \(L_{2}:\quad x=6, x \le y \le 20\) | \(P(6,y)= -3y^{2}+8y -36\) | \(P(6,y)<0\) |
| \(L_{3}:\quad 0\le x \le 6, y= 20\) | \(P(x,20)= -3x^{2}-28x -800\) | \( P(x,20)<0\) |
| \(L_{4}:\quad x=0, 0 \le y \le 20\) | \(P(0,y)= -3y^{2}+20y\) | \(P(0,\frac{10}{3}) =33.33\) |
| Critical point (1, 3): | \(P(1,3) =36\) |
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 \(P(1,3) =$ 36,000.\)