Supply and Demand problem
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)
p1, p2: 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 byelemination method:
$$-10a - 5b = -14$$
$$2 (5a + 6b = 24) $$
then we have
$$-10a - 5b = -14$$
$$10a + 12b = 48 $$
$$-------$$
$$7b = 34$$
$$b={34 \over 7}$$
$$5a + 6 \left( {34 \over 7} \right)= 24$$
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
$$det(H) = 10 \times 4 - 5 \times 5=40-25=15>0$$
$$\frac{\partial^2 P}{\partial a^2}=-10<0$$
So P has a maximum in a = 0 , b= 4
$$P = 14(0)-5(0)^2+ 24(4)-3(4)^2 - 5(0)(11)= 48$$
- ۰۴/۰۴/۱۱