Why minimum of a differentiable function over a cone is in the interior of a cone?

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Let $K \subseteq \mathbb{R}^n$ be a convex cone and $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be a differentiable function. Now consider the following optimization problem

$$ \min_{x \in K} f(x) $$

Let $x^*\in K$ be a minimizer of $f(x)$.

Show that if $x^*$ is in the interior of $K$, then $\nabla f(x^*)=0$

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Hint: one way to do it is to show that $\frac{\partial f}{\partial x_i}(\vec x^*)$ vanishes for $1\le i\le n.$ Do this by considering the quotients $\frac{f(\vec x^*+t\vec e_i )-f(\vec x^*)}{t}$ and applying a result from the calculus of one variable. The quotients are well-defined if $t$ is small enough because $\vec x^*\in \text{int} K.$