3. (10 points) $A, B, C, D$ are assigned a permutation of the numbers $10, 20, 30, 40$, such that the value of the expression $\frac{1}{\mathrm{~A}-\frac{1}{B+\frac{1}{C-\frac{1}{D}}}}$ is maximized. Then the value of $A+2B+3C+4D$ is $\qquad$

Assign $A, B, C, D$ using a permutation of the numbers $10, 20, 30, 40$ to maximize the value of the expression $\frac{1}{\mathrm{~A}-\frac{1}{B+\frac{1}{C-\frac{1}{D}}}}$. Then, the value of $A+2B+3C+4D$ is $\qquad$
【Analysis】First, observe that the numerator is 1, so the size of the fraction is determined by the denominator. Therefore, the denominator should take the smallest value. Discussing each case, we derive the values of $A, B, C, D$.

【Solution】According to the analysis, to maximize the fraction
$$
\frac{1}{\mathrm{~B}+\frac{1}{C-\frac{1}{D}}}
$$
the denominator $\mathrm{A}-\frac{1}{\mathrm{~B}+\frac{1}{\mathrm{C}-\frac{1}{\mathrm{D}}}}$ should take the smallest value.

Then $A=10$, so the fraction $\frac{1}{\mathrm{~B}+\frac{1}{\mathrm{C}-\frac{1}{\mathrm{D}}}}$ after $A$ should be maximized, which means the denominator should be minimized, so $B=20$. The fraction $\frac{1}{C-\frac{1}{\mathrm{D}}}$ added to $B$ should be minimized, which means the denominator should be as large as possible, so $C=40, D=30$. Therefore, $A+2 B+3 C+4 D=10+2 \times 20+3 \times 40+4 \times 30=290$.

Thus, the answer is: 290