Question

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Answer 1

The profit is the difference between the revenue and the cost, so we have

$P(x,y)=6x+4y-(x^2-2xy+6y^2+10x-60y-2)=-x^2+2xy-4x-6y^2+64y+2$

Let's compute the partial derivatives:

$\dfrac{\partial P}{\partial x}=-2x+2y-4,\quad \dfrac{\partial P}{\partial y}=2x-12y+64$

To find the point of maximum, we have to impose the gradient to be null:

$\begin{cases}-2x+2y-4=0\\2x-12y+64=0\end{cases}$

If we sum the two equations, we have

$-10y+60=0 \iff y=6$

And we deduce from the first equation

$-2x+2\cdot 6-4=0 \iff -2x+8=0 \iff x=4$

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

$P(4,6)=-4^2+2\cdot 4\cdot 6-4\cdot 4-6\cdot 6^2+64\cdot 6+2=186$