Maximize Q=3xy^2, where x and y are positive numbers such that x+y^2=2
1 answer:
Sub x = 2-y^2 to Q, we get: Q = 3(2-y^2)*y^2 let y^2 = k Q = 3(2-k)k = 3(2k-k^2) 2k-k^2 has a max when k = 1 Then y^2 = 1 -> y = 1 or -1
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