Check whether the function is convex or concave or neither.

Asked Jan 26 '22 at 10:22

Active Jan 26 '22 at 10:42

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So I have a given function $f:S\rightarrow\mathbb{R}$ where $S=\{x\in\mathbb{R}^2:x_1,x_2>0\}$

$f(x)=12x_1^\frac{1}{3}x_2^\frac{1}{2}$

How do I check the convexity of the function ? I thought of proceeding via finding the eigenvalues of the hessian matrix, but it seems like a tedious job, and I am not even sure if that would lead to anything conclusive. Any help is appreciated.

edited Jan 26 '22 at 10:42

asked Jan 26 '22 at 10:22

The Limit Does Not Exist

1

perhaps show your working of computing the Hessian matrix? – Siong Thye Goh Jan 26 '22 at 10:31
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Hint: Draw a picture of $f(1, x_2) = 12 \sqrt{x_2}$. Does that look convex? – Martin R Jan 26 '22 at 10:37
@MartinR Yup, I plotted the graph, looks concave, is there any clean way to prove that tho ? Or hessian is the only way to go ? – The Limit Does Not Exist Jan 26 '22 at 10:45
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Computing the Hessian should be not too bad if you extract common factors from the rows and columns of the determinant. I am not aware of a shorter method right now. – Martin R Jan 26 '22 at 10:53

Check whether the function is convex or concave or neither.

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