= Solution
{c}
Take the <semidirect product of a Lie algebra and a module>
$$
L=\mathfrak{sl}_2(k)\ltimes k^2,\qquad[(X,u),(Y,v)]=([X,Y],Xv-Yu),
$$
where $k^2$ is the <defining representation> of the <sl2 Lie algebra>. To see directly that this is a <Lie algebra>, realize $(X,u)$ as the block <matrix> $\left(\begin{smallmatrix}X&u\\0&0\end{smallmatrix}\right)$. These <matrices> are closed under the <operator commutator>, whose <Jacobi identity> holds in every <associative algebra>.
Using the standard <sl2 triple> $e,h,f$, its <Lie brackets> show $[\mathfrak{sl}_2,\mathfrak{sl}_2]=\mathfrak{sl}_2$: indeed $h=[e,f]$, $e=[h,e]/2$, and $f=-[h,f]/2$. Also $h=\operatorname{diag}(1,-1)$ is invertible on $k^2$. For any $u\in k^2$, put $v=h^{-1}u$; then
$$
[(h,0),(0,v)]=(0,u).
$$
It follows that the <derived algebra> contains both $\mathfrak{sl}_2\oplus0$ and $0\oplus k^2$, so $[L,L]=L$: this is a <perfect Lie algebra>.
However, $0\oplus k^2$ is a nonzero <ideal of a Lie algebra> with zero <Lie bracket>. It is therefore a <solvable Lie algebra> and is contained in the <solvable radical> of $L$. A <semisimple Lie algebra> has zero <solvable radical>, so
$$
\boxed{[L,L]=L,\qquad0\ne k^2\subseteq\operatorname{rad}L,\qquad\dim L=5.}
$$
This illustrates the <perfect semidirect product criterion>: the <Lie brackets> internal to the first summand generate it, while its action generates the second summand.
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