Question

company can produce and sell x-units of one commodity and y-units of other commodity

at a pro€t given as P(x, y) = 400x + 500y − x

2 − y

2 − xy − 20000. Find the units and amount

for which the pro€t is maximum and minimum.

Answer 1

$$P(x, y)=400x+500y-x^2-y^2-xy-20000$$

$$\dfrac{\partial P}{\partial x}=400-2x-y$$

$$\dfrac{\partial P}{\partial y}=500-2y-x$$

Find the critical point(s)

$$\begin{cases} \dfrac{\partial P}{\partial x}=0 \\ \\ \dfrac{\partial P}{\partial y}=0 \end{cases}=> \begin{cases} 400-2x-y=0 \\ \\ 500-2y-x=0 \end{cases}$$

$$=> \begin{cases} y=400-2x \\ \\ 500-800+4x-x=0 \end{cases}=> \begin{cases} x=100 \\ \\ y=200 \end{cases}$$

$$\dfrac{\partial^2 P}{\partial x^2}=-2$$

$$\dfrac{\partial^2 P}{\partial x \partial y}=\dfrac{\partial^2 P}{\partial y \partial x}=-1$$

$$\dfrac{\partial^2 P}{\partial y^2}=-2$$

$$D(x, y)=\begin{vmatrix} -2 & -1 \\ -1 & -2 \end{vmatrix}=4-1=3$$

$$\dfrac{\partial^2 P}{\partial x^2}(100,200)=-2<0$$

$$D(100, 200)=3>0$$

It can be stated that $(100,200)$ is a relative maximum.

$$P(100,200)=400(100)+500(200)$$

$$-(100)^2-(200)^2-100(200)-20000=50000$$

The Profit has the maximum with value of $50000$ at $x=100$ units and $y=200$ units.

The Profit has no minimum.