One of the most important functions we will encounter in this book is the Cobb-Douglas function: \\[ y=\left(x_{1}\right)^{\alpha}\left(x_{2}\right)^{\beta} \\] where \(\alpha\) and \(\beta\) are positive constants that are each less than 1 a. Show that this function is quasi-concave using a "brute force" method by applying Equation 2.114 b. Show that the Cobb-Douglas function is quasi-concave by showing that any contour line of the form \(y=c\) (where \(c\) is any positive constant is convex and therefore that the set of points for which \(y>c\) is a convex set. c. Show that if \(\alpha+\beta>1\) then the Cobb-Douglas function is not concave (thereby illustrating again that not all quasiconcave functions are concave). Note: The Cobb-Douglas function is discussed further in the Extensions to this chapter.

To show that the Cobb-Douglas function is quasi-concave, we can use the definition of a quasi-concave function given by Equation 2.114, which states that if for any two points \(x_1, x_2 \in X\) and any \(\lambda \in [0, 1]\), the following inequality holds: \\[ f(\lambda x_1 + (1-\lambda) x_2) \geq \min[f(x_1), f(x_2)] \\] We have the Cobb-Douglas function given by \(y=f(x_1,x_2)=(x_1)^{\alpha}(x_2)^{\beta}\), where \(\alpha\) and \(\beta\) are positive constants that are each less than 1. Let's choose two points \(x_1 = (a_1, a_2)\) and \(x_2=(b_1, b_2)\) and consider \(\lambda \in [0, 1]\). We then have: \(f(x_1) = (a_1)^{\alpha}(a_2)^{\beta}\) and \(f(x_2) = (b_1)^{\alpha}(b_2)^{\beta}\) Now, we need to evaluate the function at the point given by \(\lambda x_1 + (1-\lambda)x_2=(a_1\lambda+b_1(1-\lambda),a_2\lambda+b_2(1-\lambda))\). We have: \(f(\lambda x_1+(1-\lambda) x_2)=((a_1\lambda+b_1(1-\lambda))^{\alpha}((a_2\lambda+b_2(1-\lambda))^{\beta}\). To satisfy the inequality of quasi-concave functions, we need to prove that: \(((a_1\lambda+b_1(1-\lambda))^{\alpha}((a_2\lambda+b_2(1-\lambda))^{\beta} \geq \min[f(x_1), f(x_2)]\) Using the fact that \(A^{\lambda}B^{1-\lambda} \geq \lambda A + (1-\lambda)B\) for positive A, B and \(\lambda \in [0, 1]\) (which can be proved using Jensen's inequality), we get: \(((a_1\lambda+b_1(1-\lambda))^{\alpha}((a_2\lambda+b_2(1-\lambda))^{\beta} \geq (\lambda a_1^{\alpha} + (1-\lambda) b_1^{\alpha})(\lambda a_2^{\beta} + (1-\lambda) b_2^{\beta})\) Now, notice that: \((\lambda a_1^{\alpha} + (1-\lambda) b_1^{\alpha})(\lambda a_2^{\beta} + (1-\lambda) b_2^{\beta}) \geq \min[f(x_1), f(x_2)]\) Therefore, the Cobb-Douglas function is quasi-concave.

This approach requires showing that the contour lines of the Cobb-Douglas function, given by \(y=c\), are convex and the set of points for which \(y>c\) is a convex set. Let's define the contour line of the Cobb-Douglas function as: \(y = (x_1)^{\alpha}(x_2)^{\beta} = c\) To show that this contour line is convex, we can use the Hessian matrix to check the property of the function. The Hessian matrix is given by: \(H = \begin{bmatrix} \frac{\partial^2y}{\partial x_1^2} & \frac{\partial^2y}{\partial x_1\partial x_2} \\ \frac{\partial^2y}{\partial x_2\partial x_1} & \frac{\partial^2y}{\partial x_2^2} \end{bmatrix}\) Taking the necessary second-order partial derivatives, we get: \(\frac{\partial^2y}{\partial x_1^2} = \alpha(\alpha-1)(x_1)^{\alpha-2}(x_2)^{\beta}\), \(\frac{\partial^2y}{\partial x_2^2} = \beta(\beta-1)(x_1)^{\alpha}(x_2)^{\beta-2}\), \(\frac{\partial^2y}{\partial x_1\partial x_2}=\frac{\partial^2y}{\partial x_2\partial x_1}=\alpha\beta(x_1)^{\alpha-1}(x_2)^{\beta-1}\). Thus, the Hessian matrix is: \(H = \begin{bmatrix} \alpha(\alpha-1)(x_1)^{\alpha-2}(x_2)^{\beta} & \alpha\beta(x_1)^{\alpha-1}(x_2)^{\beta-1} \\ \alpha\beta(x_1)^{\alpha-1}(x_2)^{\beta-1} & \beta(\beta-1)(x_1)^{\alpha}(x_2)^{\beta-2} \end{bmatrix}\) The function is convex if the Hessian matrix is positive semidefinite, meaning that its determinant is non-negative: \(\det(H) = \alpha^2\beta^2(x_1)^{2\alpha-2}(x_2)^{2\beta-2}-\alpha\beta(\alpha-1)(\beta-1)(x_1)^{2\alpha-2}(x_2)^{2\beta-2} \geq 0\) Notice that \(\alpha\beta(\alpha-1)(\beta-1) \leq \alpha^2\beta^2\) because \(\alpha, \beta < 1\), so the determinant of the Hessian is non-negative, and therefore, the Cobb-Douglas function is quasi-concave.

A function is concave if the negative of the function is convex. Let's check the conditions for the Cobb-Douglas function to be concave by considering the negative of this function: \(-y = -(x_1)^{\alpha}(x_2)^{\beta}\) Using the Hessian matrix of the negative function, we have: \(\frac{\partial^2(-y)}{\partial x_1^2} = \alpha(\alpha-1)(x_1)^{\alpha-2}(x_2)^{\beta}\), \(\frac{\partial^2(-y)}{\partial x_2^2} = \beta(\beta-1)(x_1)^{\alpha}(x_2)^{\beta-2}\), \(\frac{\partial^2(-y)}{\partial x_1\partial x_2}=\frac{\partial^2(-y)}{\partial x_2\partial x_1}=-\alpha\beta(x_1)^{\alpha-1}(x_2)^{\beta-1}\). The determinant of the Hessian matrix for the negative function is: \(\det(-H) = -\alpha^2\beta^2(x_1)^{2\alpha-2}(x_2)^{2\beta-2}+\alpha\beta(\alpha-1)(\beta-1)(x_1)^{2\alpha-2}(x_2)^{2\beta-2}\) Now, if \(\alpha+\beta>1\), then \(\alpha-1\) and \(\beta-1\) are positive values. Therefore, \(\det(-H) > 0\) for \(\alpha+\beta>1\), which means that the negative of the Cobb-Douglas function is not convex, and therefore, the original function is not concave under this condition.