Suppose a company produces two products A and B which have demand functions \begin{gather*} D_{A}=30-P_{A} \\ D_{B}=25-P_{B} \end{gather*} With $P_{A}$ and $P_{B}$ being their prices. If the combined cost function is \begin{gather*} C=x^{2}+2xy+y^{2}+10 \end{gather*} where $x$ denotes number of units of product A and $y$ denotes number of units of product B, what is the profit-maximizing output?
My attempt at this so far, has been as follows:
Profit = Revenue - Costs
Where Revenue is given by $P_{A}x+P_{B}y = x(30-D_{A})+y(25-D_{B})$
Thus, we have \begin{gather*} Profit = P = x(30-D_{A})+y(25-D_{B})-(x^{2}+2xy+y^{2}+10) \\ P=30x-D_{A}x-x^{2}+25y-D_{B}y-y^{2}-2xy-10 \end{gather*}
So this is the equation that we need to maximise. To do this, I assume I must differentiate the expression for $P$ (implicitly) to find its critical points. Is this correct? If so, how would I differentiate the left hand side of the equation? Would it just be $0$?
There is no implicit differentiation necessary. First you have to consider that $D_A$ and $D_B$ are the quantity units x and y respectively. The profit function becomes:
$P(x,y)=30x-x^2-x^{2}+25y-y^2-y^{2}-2xy-10$
$=30x-2x^2+25y-2y^2-2xy-10$
Differenting w.r.t $x$ and $y$
$\frac{\partial P}{\partial x}=30-4x-2y=0$
$\frac{\partial P}{\partial y}=25-4y-2x=0$
Solving this little equation system leads to the solution for $x$ and $y$. Because of the negative coefficients of $x^2$ and $y^2$ this stationary point will be the maximum.