One of the most important functions we will encounter in this book is the Cobb-Douglas function: \\[ y=(x \\] where a and \(P\) are positive constants that are each less than one. a. Show that this function is quasi-concave using a "brute force" method by applying Equation 2.107 b. Show that the Cobb-Douglas function is quasi-concave by showing that the 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 \(a+y 8>1\) then the Cobb-Douglas function is not concave (thereby illus trating that not all quasi-concave functions are concave). (Note: The Cobb-Douglas function is discussed further in the Extensions to this chapter.)

To check for quasi-concavity using the Hessian Matrix, first, we need to find the matrix of second derivatives for the given function. Start by finding the first partial derivatives with respect to \(x_1\) and \(x_2\). First derivative w.r.t. \(x_1\): \\[f_{x_1} = \alpha x_1^{\alpha-1} x_2^{\beta}\\] First derivative w.r.t. \(x_2\): \\[f_{x_2} = \beta x_1^{\alpha} x_2^{\beta-1}\\] Next, find the second partial derivatives: Second derivative w.r.t. \(x_1\): \\[f_{x_1x_1} = \alpha (\alpha - 1) x_1^{\alpha-2} x_2^{\beta}\\] Second derivative w.r.t. \(x_2\): \\[f_{x_2x_2} = \beta (\beta - 1) x_1^{\alpha} x_2^{\beta-2}\\] Second derivative w.r.t. both \(x_1\) and \(x_2\): \\[f_{x_1x_2} = f_{x_2x_1} = \alpha\beta x_1^{\alpha-1} x_2^{\beta-1}\\] Now, form the Hessian matrix \\[ H = \begin{pmatrix} f_{x_1x_1} & f_{x_1x_2} \\ f_{x_2x_1} & f_{x_2x_2} \end{pmatrix} \\] A function is quasi-concave if its Hessian matrix is negative semi-definite. To check negative semi-definiteness, we will check whether the determinant of Hessian is non-negative and if at least one of the diagonal entries is non-positive. Determinant: \\[\text{det}(H) = f_{x_1x_1}f_{x_2x_2} - f_{x_1x_2}^2 = \alpha(\alpha-1)\beta(\beta-1)x_1^{\alpha-2}x_2^{\beta-2}-\alpha^2\beta^2x_1^{\alpha+\beta-2}x_2^{\alpha+\beta-2}\\] Notice that since \(\alpha\) and \(\beta\) are positive constants less than 1, then \(\alpha-1 < 0\) and \(\beta-1 < 0\). This means that \(\alpha(\alpha-1) < 0\) and \(\beta(\beta-1) < 0\) and therefore, the determinant is non-negative. Taking a look at the diagonal elements, for example \(f_{x_1x_1}\), since \(\alpha(\alpha - 1)\) is non-positive, we have negative semi-definite Hessian, which implies that the Cobb-Douglas function is quasi-concave.

We now want to show that any contour line of the form \(y = c\), where \(c\) is a positive constant, is convex. Rewrite the Cobb-Douglas function \(y\) in terms of contour lines: \\[ c = x_1^{\alpha}x_2^{\beta} \\] Now, consider two points \((x_1^1, x_2^1)\) and \((x_1^2, x_2^2)\), let \(\lambda \in (0,1)\) and form the convex combination of these points, with \(x_1^* = \lambda x_1^1 + (1-\lambda) x_1^2\) and \(x_2^* = \lambda x_2^1 + (1-\lambda) x_2^2\). Our goal is to show that \(c^* \ge c\), where \(c^* = (x_1^*)^{\alpha}(x_2^*)^{\beta}\), because this means that the contour lines are convex. Using the convex combination property, calculate \(c^*\): \\[ c^* = [\lambda x_1^1 + (1-\lambda) x_1^2]^{\alpha}[\lambda x_2^1 + (1-\lambda) x_2^2]^{\beta} \\] Use Jensen's inequality which states that if $$\phi$$ is a concave function, then \(\phi(\lambda x_1^1 + (1-\lambda) x_1^2) \ge \lambda\phi(x_1^1) +(1-\lambda) \phi(x_1^2)\). Since \(\alpha\) and \(\beta\) are less than 1, we know that the power functions \(x^{\alpha}\) and \(x^{\beta}\) are concave. Apply Jensen's inequality separately to \(x^{\alpha}\) and \(x^{\beta}\) terms: \\[ c^* \ge [\lambda x_1^{\alpha} + (1-\lambda) x_1^{\alpha}][\lambda x_2^{\beta} + (1-\lambda) x_2^{\beta}] = c \\] Since \(c^* \ge c\), the contour lines are convex, and thus the Cobb-Douglas function is quasi-concave.

To prove that the Cobb-Douglas function is not concave when \(\alpha + \beta > 1\), we have to show that for some \(x_1^1, x_2^1, x_1^2, x_2^2\) and \(\lambda \in (0,1)\), the inequality \(f(\lambda x_1^1 + (1-\lambda) x_1^2, \lambda x_2^1 + (1-\lambda) x_2^2) < \lambda f(x_1^1, x_2^1) + (1-\lambda) f(x_1^2, x_2^2)\) holds. Consider two points: \((x_1^1, x_2^1) = (1,1)\) and \((x_1^2, x_2^2) = (0,0)\), and let \(\lambda = 1/2\). Then: \\[ f\left(\frac{1}{2}, \frac{1}{2}\right) = \left(\frac{1}{2}\right)^{\alpha}\left(\frac{1}{2}\right)^{\beta} = \left(\frac{1}{2}\right)^{\alpha + \beta} \\] \\[ \frac{1}{2}f(1, 1) + \frac{1}{2}f(0, 0) = \frac{1}{2} \\] If \(\alpha + \beta > 1\), then \(\left(\frac{1}{2}\right)^{\alpha + \beta} < \frac{1}{2}\). As a result, the function is not concave in this case.