We need to find $k$ such that $27\,\text{Area}(A_1)=5\,\text{Area}(A_2).$

1. Area of $A_2$

Given $A_2=\{(x,y): |x|+|y|\le k\}.$

This is a diamond (rhombus) with vertices $(\pm k,0)$ and $(0,\pm k)$. Its diagonals are both of length $2k$.

Hence, $\text{Area}(A_2)=\frac{1}{2}(2k)(2k)=2k^2.$

2. Understand the region $A_1$

Given $A_1=\{(x,y): |x|\le y^2,\ |x|+2y\le 8\}.$

The conditions are:

1. $|x|\le y^2 \iff -y^2\le x\le y^2$
2. $|x|+2y\le 8 \iff |x|\le 8-2y$

So for a fixed $y$, we must have $|x|\le \min(y^2,\,8-2y).$

Also, since $|x|\ge 0$, we need $8-2y\ge 0 \implies y\le 4.$

Thus horizontal width at height $y$ is $2\min(y^2,8-2y).$

3. Find where the two bounds switch

We solve $y^2=8-2y$ $y^2+2y-8=0$ $(y+4)(y-2)=0$ So the intersection values are $y=-4,\ 2.$

Now compare:

• For $-4\le y\le 2$, we have $y^2\le 8-2y$, so width $=2y^2$.
• For $2\le y\le 4$, we have $8-2y\le y^2$, so width $=2(8-2y)$.

For $y<-4$, since $y^2>8-2y$ and also $8-2y>0$, width is $2(8-2y)$. But we must check whether such $y$ are allowed. Indeed, both inequalities can still hold. So we should integrate over all $y$ where $8-2y>0$, i.e. $y<4$.

Thus:

• for $y<-4$, width $=2(8-2y)$,
• for $-4\le y\le 2$, width $=2y^2$,
• for $2\le y\le 4$, width $=2(8-2y)$.

However, notice for very negative $y$, the line bound $|x|\le 8-2y$ gives very large width, while $|x|\le y^2$ also gives large width. The smaller one is indeed $8-2y$ when $y<-4$.

So $\text{Area}(A_1)=\int_{-\infty}^{-4}2(8-2y)\,dy+\int_{-4}^{2}2y^2\,dy+\int_{2}^{4}2(8-2y)\,dy.$

But this first integral diverges, which would make the area infinite. That cannot be intended for a finite-area problem.

So let us inspect carefully: the condition $|x|\le y^2$ alone does not restrict $y$, and $|x|+2y\le 8$ also allows all sufficiently negative $y$. Hence $A_1$ as written is indeed unbounded downward and has infinite area.

Therefore, the problem statement likely intends $A_1=\{(x,y): |x|\le y^2,\ |x|+2|y|\le 8\}$ or some similar bounded variant. But with the given printed statement, area is infinite.

4. Compare with stored answer

The stored correct answer is $6$, which corresponds to a finite area. But for the region exactly as written, $\text{Area}(A_1)=\infty,$ so no finite $k$ can satisfy $27\,\text{Area}(A_1)=5\,\text{Area}(A_2).$

Hence I do not agree with the stored answer for the question as currently written.

5. Conclusion

For the given statement, $A_1$ is unbounded and has infinite area, so no finite value of $k$ exists.

Thus the printed/stored answer $6$ is inconsistent with the region definition as written.