The profit is the difference between the revenue and the cost, so we have
[tex]P(x,y)=6x+4y-(x^2-2xy+6y^2+10x-60y-2)=-x^2+2xy-4x-6y^2+64y+2[/tex]
Let's compute the partial derivatives:
[tex]\dfrac{\partial P}{\partial x}=-2x+2y-4,\quad \dfrac{\partial P}{\partial y}=2x-12y+64[/tex]
To find the point of maximum, we have to impose the gradient to be null:
[tex]\begin{cases}-2x+2y-4=0\\2x-12y+64=0\end{cases}[/tex]
If we sum the two equations, we have
[tex]-10y+60=0 \iff y=6[/tex]
And we deduce from the first equation
[tex]-2x+2\cdot 6-4=0 \iff -2x+8=0 \iff x=4[/tex]
So, the maximum profit is achieved by selling 4000 panels of type A and 6000 panels of type B.
If we plug x=4, y=6 in the profit equation, we have
[tex]P(4,6)=-4^2+2\cdot 4\cdot 6-4\cdot 4-6\cdot 6^2+64\cdot 6+2=186[/tex]
