Question

exercise 4. (2 p)
a company produces two different types of products (a and b). the daily cost of producing x units of a
and y units of b is ( c(x,y) = 0.04x^2 + 0.01xy + 0.01y^2 + 4x + 2y + 500 ). they sell product a for 15 euros per unit
and product b for 9 euros per unit. study the maximum profit.

Explanation:

Step1: Define profit function

Profit = Revenue - Cost.
Revenue: $R(x,y) = 15x + 9y$
Cost: $C(x,y) = 0.04x^2 + 0.01xy + 0.01y^2 + 4x + 2y + 500$
Profit:

$$\begin{align*} P(x,y)&=R(x,y)-C(x,y)\\ &=15x + 9y - (0.04x^2 + 0.01xy + 0.01y^2 + 4x + 2y + 500)\\ &=-0.04x^2 -0.01xy -0.01y^2 +11x +7y -500 \end{align*}$$

Step2: Find partial derivatives

Compute $P_x$ and $P_y$, set to 0.
$P_x = -0.08x -0.01y +11 = 0$
$P_y = -0.01x -0.02y +7 = 0$
Rewrite as system:

$$\begin{cases} 0.08x + 0.01y = 11 \\ 0.01x + 0.02y = 7 \end{cases}$$

Multiply first eq by 100: $8x + y = 1100$
Multiply second eq by 100: $x + 2y = 700$

Step3: Solve the system

From first eq: $y = 1100 - 8x$
Substitute into second eq:

$$ x + 2(1100 - 8x) = 700 $$

$$ x + 2200 - 16x = 700 $$

$$ -15x = 700 - 2200 = -1500 $$

$$ x = 100 $$

Substitute $x=100$ into $y=1100-8x$:

$$ y = 1100 - 8(100) = 300 $$

Step4: Second derivative test

Compute second partials:
$P_{xx} = -0.08$, $P_{yy} = -0.02$, $P_{xy} = -0.01$
Hessian determinant:

$$ D = P_{xx}P_{yy} - (P_{xy})^2 = (-0.08)(-0.02) - (-0.01)^2 = 0.0016 - 0.0001 = 0.0015 $$

Since $D>0$ and $P_{xx}<0$, $(100,300)$ is a maximum.

Step5: Calculate max profit

Substitute $x=100,y=300$ into $P(x,y)$:

$$\begin{align*} P(100,300)&=-0.04(100)^2 -0.01(100)(300) -0.01(300)^2 +11(100)+7(300)-500\\ &=-400 -300 -900 +1100 +2100 -500\\ &=1100 \end{align*}$$

Answer:

The maximum profit is 1100 euros, achieved when producing 100 units of product A and 300 units of product B.