Buy 56$ 7 toys type A. Then we will earn 2*7=14 $. If we buy type B, we will earn 3*4=12$. from this it seems that it is profitable to sell toys A. But here we must take into account that no more than 2000 toys of two types will be sold in total, and not only the costs are important, but also the individual profitability. I.e. we have the following system of inequations:
"\\begin{Bmatrix}\n 8A+14B\u226420000\\\\\n A+B\u22642000\n\\end{Bmatrix}\n\\iff\\begin{Bmatrix}\n 8(A+B)+6B\u226420000 \\\\\n A+B\u22642000\n\\end{Bmatrix}"
(let me remind you that we strive to maximize 2A+3B)
2A+3B=2*(A+B)+B≤2*2000+B=4000+B
This means that we want to buy type b toys.
This means that we can already write the system as
"\\begin{Bmatrix}\n 16000+6B\u226420000 \\\\\n A+B=2000\n\\end{Bmatrix}\\iff\\begin{Bmatrix}\n 3B\u22642000 \\\\\n A+B=2000\n\\end{Bmatrix}"
A=1334
B=666
2A+3B=2668+1998=4666
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