= Solution
Every finite-dimensional irreducible <Lie algebra representation> has <highest weight> $\lambda=a\Lambda_1+b\Lambda_2$ for unique integers $a,b\ge0$. The <Weyl vector> is
$$
\rho=\frac12\big((\varepsilon_1-\varepsilon_2)+\varepsilon_2+\varepsilon_1+(\varepsilon_1+\varepsilon_2)\big)
=\frac32\varepsilon_1+\frac12\varepsilon_2,
$$
so $\lambda+\rho=(a+b/2+3/2,\,b/2+1/2)$. In the <Weyl dimension formula>, the factors for the four positive roots, ordered as $\varepsilon_1-\varepsilon_2,\varepsilon_2,\varepsilon_1,\varepsilon_1+\varepsilon_2$, are respectively
$$
a+1,\qquad b+1,\qquad\frac{2a+b+3}{3},\qquad\frac{a+b+2}{2}.
$$
Hence
$$
\boxed{\dim L(a\Lambda_1+b\Lambda_2)=\frac{(a+1)(b+1)(a+b+2)(2a+b+3)}6.}
$$
In particular the two fundamental representations have dimensions five and four. The highest root is $\theta=\varepsilon_1+\varepsilon_2=2\Lambda_2$, so the <Adjoint representation> has dimension ten. These also fix the dimensions of the crystals below. Our numbering places the long root first; interchanging a B2/C2 numbering without interchanging the associated fundamental-weight labels would give the wrong dimension formula.
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