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The function is $f(K,L)= AK^{a}L^{b}$ on the set of points $(K,L)$ with $K\geq 0$ and $L\geq 0$, assuming $A>0$ How do I find the Hessian and check for concavity?

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    $\begingroup$ Do you know how to take a derivative? $\endgroup$ – Herr K. Aug 15 at 16:26
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As @Herr K. stated, the beginning point is being able to take a derivative. The Hessian is a matrix equivalent to a second order derivative sometimes denoted as $\nabla^{2}$. Start by finding the gradient, $\nabla$ which is a vector of first order derivatives of every variable in the function. In your case, this would be
$$ \nabla= \left[ {\begin{array}{cc} \frac{\partial{f}}{\partial{K}} \\ \frac{\partial{f}}{\partial{L}} \\ \end{array} } \right] = \left[ {\begin{array}{cc} aAK^{a-1}L^{b} \\ bAK^{a}L^{b-1} \\ \end{array} } \right] $$ The Hessian, would be the next (second) derivative of each of the above with respect to both variables and it takes the form: $$ \nabla^{2}= \left[ {\begin{array}{cc} \frac{\partial^{2}f}{\partial{K}^2} & \frac{\partial^{2}f}{\partial{K}\partial{L}} \\ \frac{\partial^{2}f}{\partial{L}\partial{K}} & \frac{\partial^{2}f}{\partial{L}^2} \\ \end{array} } \right] $$ Using whatever this would be in your case (left to you to compute) you can take the determinant of the resulting matrix and, with values for your parameters $a$ and $b$ (or at least their signs like with the other variables and parameters) you can determine whether the determinant is positive or negative and determine the concavity. If it is positive definite or positive semi-definite, this would imply either strict or non-strict convexity, respectively. If you don't know how to take a determinant it is simply dont by using a matrix say

$$ A= \left[ {\begin{array}{cc} a_{1} & a_{2} \\ a_{3} & a_{4} \\ \end{array} } \right] $$ and then taking its determinant using the operation (without explanation for sake of brevity): $$ A= {\begin{vmatrix} a_{1} & a_{2} \\ a_{3} & a_{4} \\ \end{vmatrix} } = a_{1}a_{4}-a_{2}a_{3} $$ Using this with the above, you should be able to figure the rest out and figure out whether or not its concave! (hint: for $a,b\geq 0$ and $a+b\leq1$ it is concave)

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  • $\begingroup$ thank you for your help, sir. $\endgroup$ – Grangerluck Aug 15 at 20:56
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    $\begingroup$ Ofcourse! Could you accept the answer? It helps get the site off of Beta :) $\endgroup$ – Brennan Aug 16 at 15:18
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    $\begingroup$ " for a,b≤0 it is concave" This is wrong. a ≥ 0, b ≥ 0, and a + b ≤ 1 are required for the function to be concave on the stated domain( "the set of points $(K,L)$ with $K≥0$ and $L≥0$") $\endgroup$ – user18214 Aug 16 at 16:21
  • $\begingroup$ Oh oops my mistake, thank you for pointing that out. $\endgroup$ – Brennan Aug 16 at 21:43

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