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Question
If x + y = 3 show that the maximum value of x2y is 4.
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Solution
x + y = 3
∴ y = 3 – x
Let T = x2y = x2(3 – x) = 3x2 – x3
Differentiating w.r.t. x, we get
`"dT"/("d"x) = 6"x" - 3"x"^2` ....(i)
Again, differentiating w.r.t. x, we get
`("d"^2"T")/("d"x^2) = 6 - 6"x"` ...(ii)
Consider, `"dT"/("d"x) = 0`
∴ 6x – 3x2 = 0
∴ x = 2
For x = 2,
`(("d"^2"T")/"dx"^2)_(x = 2)` = 6 – 6(2)
= 6 – 12
= – 6 < 0
Thus, T, i.e., x2y is maximum at x = 2
For x = 2, y = 3 – x = 3 – 2 = 1
∴ Maximum value of T = x2y = (2)2(1) = 4
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