فیزتک

A certain company offers two products (a, b, measured in thousands of units). Their respective demand equations are: \(4a+p_1=14\) and \(2b + p_2 = 24\) where \(p_1\), \(p_2\) stand for respectively- the selling price per unit of each product. The overall cost is given by \(a^ 2 + 5ab + b ^ 2\) Study the maximum profit. (2.5 points)

To study the maximum profit for the given company, we need to:

1. Define the profit function

a, b: quantities sold (in thousands)

\(p_1, p_2\): prices per unit of products a and b

2. From demand equation s, solve for price as a function of quantity:

Given:

$$ 4a + p_1 = 14 \rightarrow p_1 = 14-4a$$

$$ 2b + p_2 = 24 \rightarrow p_2 = 24-2b$$

3. Revenue function

Revenue = (price of a × quantity a) + (price of b × quantity b)

$$R(a,b) = p_1 \times a + p_2 \times b$$

4. Cost function:

$$a^ 2 + 5ab + b ^ 2$$

5. Profit function = Revenue - Cost

$$P(a,b) = R(a,b) - C(a,b)$$

$$P(a,b) = a p_1 + b p_2 - (a^ 2 + 5ab + b ^ 2)$$

$$P = a (14-4a)+ b (24-2b)- (a^ 2 + 5ab + b ^ 2)$$

$$P = 14a-4a^2+ 24b-2b^2- a^ 2 - 5ab - b ^ 2$$

$$P = 14a-5a^2+ 24b-3b^2 - 5ab$$

6. Maximize profit function

Take partial derivatives and set them to 0:

$$\frac{\partial P}{\partial a}=0 \rightarrow 14 - 10a - 5b = 0$$

$$\frac{\partial P}{\partial b}=0 \rightarrow 24 - 6b - 5a = 0$$

Now solve equations by elimination method:

\begin{cases}-10a - 5b = -14 \\2 (5a + 6b = 24) \end{cases}

then we have

\begin{cases}-10a - 5b = -14 \\10a + 12b = 48  \end {cases}

$$7b = 34$$

$$b={34 \over 7}$$

$$5a + 6 \left( {34 \over 7} \right)= 24$$

multiply both side by 7

$$35a + 204 = 168$$

$$35a = 168-204$$

$$35a = -36$$

$$a = -{36 \over 35}$$

8. Conclusion

The critical point is at

$$a = -{36 \over 35}, b={34 \over 7}$$

but a<0, which is not valid scince quantity sold cannot be negative.

so $$a = 0 ; b = 4$$

second derivative

$$\frac{\partial^2 P}{\partial a^2}=-10;\frac{\partial^2 P}{\partial b^2}=-4;\frac{\partial^2 P}{\partial ab}=-5$$

$$H = \begin{pmatrix} \frac{\partial^2 P}{\partial a^2} & \frac{\partial^2 P}{\partial ab} \\\ \frac{\partial^2 P}{\partial ab} & \frac{\partial^2 P}{\partial b^2} \end{pmatrix}$$

$$H = \begin{pmatrix}-10 & -5\\\ -5 & -4\end{pmatrix}$$

calculate determination of Hessian matrix

$$|H| = 10 \times 4 - 5 \times 5=40-25=15>0$$

$$\frac{\partial^2 P}{\partial a^2}=-10<0$$

So P has a maximum in \(a = 0 , b= 4\)

$$P = 14(0)-5(0)^2+ 24(4)-3(4)^2 - 5(0)(11)= 48$$