Let $$A_1 = \{(x,y) : |x| \leq y^2, |x| + 2y \leq 8\}$$ and $$A_2 = \{(x,y) : |x| + |y| \leq k\}$$. If $$27$$ (Area $$A_1$$) $$= 5$$ (Area $$A_2$$), then $$k$$ is equal to ______

Given,

$$A_1=\{(x,y):|x|\le y^2,\ |x|+2y\le8\}$$

and

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

The region $$A_1$$ is symmetric about the $$y$$-axis.

The curves

$$|x|=y^2$$

and

$$|x|=8-2y$$

intersect at

$$y^2=8-2y$$

$$y^2+2y-8=0$$

$$y=2,-4$$

Taking the bounded upper region,

$$0\le y\le4$$

Now,

for

$$0\le y\le2,\qquad |x|\le y^2$$

and for

$$2\le y\le4,\qquad |x|\le8-2y$$

Hence,

$$\text{Area}(A_1) = 2\left(\int_0^2 y^2\,dy+\int_2^4(8-2y)\,dy\right) $$

$$ = 2\left(\frac83+4\right) = \frac{40}{3}$$

Now,

$$|x|+|y|\le k$$

represents a square with diagonals along the coordinate axes.

Its area is

$$\text{Area}(A_2)=2k^2$$

Given,

$$27(\text{Area }A_1)=5(\text{Area }A_2)$$

$$27\cdot\frac{40}{3}=5\cdot2k^2$$

$$360=10k^2$$

$$k^2=36$$

$$k=6$$

Hence,

$$\boxed{6}$$