Let A_1=(x,y):|x|≤ y^2,|x|+2y≤ 8 and A_2=(x,y):|x|+|y|≤ k. If 27 (Area A_1) =5(Area A_2), then k is equal to

${A}_{1}={(x,y):|x|\leq {y}^{2},|x|+2y\leq 8}$A1​=(x,y):∣x∣≤y2,∣x∣+2y≤8 and

$${A}_{2}={(x,y):|x|+|y|\leq k}$$A2​=(x,y):∣x∣+∣y∣≤k

Area $\displaystyle ({A}_{1})=2[{\int }_{0}^{2}{y}^{2}dy+{\int }_{2}^{4}(8-2y)dy]$(A1​)=2[∫02​y2dy+∫24​(8−2y)dy]

$$=2[{(\frac{{y}^{3}}{3})}_{0}^{2}+{(8y-{y}^{2})}_{2}^{4}]$$=2[(3y3​)02​+(8y−y2)24​]

Area $({A}_{1})=2\times \frac{20}{3}=\frac{40}{3}$(A1​)=2×320​=340​

Area $({A}_{2})=4\times \frac{1}{2}{k}^{2}$(A2​)=4×21​k2

Area $({A}_{2})=2{k}^{2}$(A2​)=2k2

Given $27(\mathrm{Area}{A}_{1})=5(\mathrm{Area}{A}_{2})$27(AreaA1​)=5(AreaA2​)

$$\Rightarrow 27\times \frac{40}{3}=5\times 2{k}^{2}$$⇒27×340​=5×2k2

$$\Rightarrow k=6$$⇒k=6