Question

Let $A_{1}=\left\{(x, y):|x| \leq y^{2},|x|+2 y \leq 8\right\}$ and $A_{2}=\{(x, y):|x|+|y| \leq k\}$. If 27 (Area $\left.A _{1}\right)=5$ (Area $A _{2}$ ), then $k$ is equal to:

Answer 1

$A_{1}=\left\{(x, y):|x| \leq y^{2},|x|+2 y \leq 8\right\}$ and
$A_{2}=\{(x, y):|x|+|y| \leq k\}$

$\text{area}\left(A_{1}\right)=2\left[\int\limits_{0}^{2} y^{2} d y+\int\limits_{2}^{4}(8-2 y) d y\right]$
$=2\left[\left(\frac{y^{3}}{3}\right)_{0}^{2}+\left(8 y-y^{2}\right)_{2}^{4}\right]$

$\text{area}\left(A_{1}\right)=2 \times \frac{20}{3}=\frac{40}{3}$
$\text{Area}\left( A _{2}\right)=4 \times \frac{1}{2} k^{2}$
Area $\left( A _{2}\right)=2 k^{2}$
Now
$27\left(\right.$ Area $\left.A _{1}\right)=5\left(\right.$ Area $\left.A _{2}\right)$
$9 \times 4=k^{2}$
$k =6$