Suppose the $k$-algebra $R$ is finite-dimensional as a vector space over $k$, for example, when $R = k[x]/<f(x)>$, where $f$ is any nonzero polynomial in $k[x]$. Then in particular $R$ is a finitely generated $k$-algebra since a vector space basis also generates $R$ as a ring. In this case since ideals are also $k$-subspaces any ascending or descending chain of ideals has at most $\mathbf{dim_k R + 1}$ distinct terms, hence $R$ satisfies both A.C.C and D.C.C- on ideals.
Doubt: How will any ideal have at most $dim_k R +1$ distinct terms? Shouldn't it be just $dim_k R$ and not $dim_k R +1$?
Suppose $f(x)=a_nx^n+\cdots+a_0$, then $R=\{b_0+b_1x+\cdots+b_{n-1}x^{n-1}+<f(x)>| \; b_0,\cdots,b_{n-1} \in k \}$ is an $\textbf{$n-$dimensional vector space}$ over $k$. So, any ideal might contain constant, linear etc terms which can be at most $n$.
In fact, for any $n$ dimensional vector space, it should be $n$ only.
P.S.: This is from Dummit and Foote, Abstract Algebra.
if you agree that the ideals are the $k$-subspaces in an $n$-dimensional space, then the number of subspaces, i.e., $d$-dimensional subspace can take the value $d=0,\dots,n$. Thus, it is $n+1$.