The area enclosed by the curves xy+4y=16 and x+y=6 is equal to :
(A)28−30loge2
(B)30−28loge2
(C)30−32loge2✓ Correct
(D)32−30loge2
Answer: (C)
Solution
xy+4y=16⇒y=x+416; x+y=6⇒y=6−x. Intersection: x+416=6−x⇒16=24+2x−x2⇒x2−2x−8=0⇒x=4,−2. At x=0, the line gives y=6 and the curve gives y=4, so the line lies above the curve on [−2,4].