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The area enclosed by the curves xy+4y=16 and x+y=6 is equal to :
(3) 30-32 e2
y= 16x+4 (rectangular hyperbola) and y=6-x
Meet: (6-x)(x+4)=16 x2-2x-8=0 x=-2, 4
[figure]
Area = -24 (6-x- 16x+4 )dx
= [6x- x22 ]-24-16 [ e(x+4) ]-24
=(16+14)-16 e4=30-32 e2
Answer: (3)
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