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If x + y = 3 show that the maximum value of x^{2}y is 4.

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#### Solution

x + y = 3

∴ y = 3 – x

Let T = x^{2}y = x^{2}(3 – x) = 3x^{2} – x^{3}^{ }

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 – 3x^{2} = 0

∴ x = 2

For x = 2,

`(("d"^2"T")/"dx"^2)_(x = 2)` = 6 – 6(2)

= 6 – 12

= – 6 < 0

Thus, T, i.e., x^{2}y is maximum at x = 2

For x = 2, y = 3 – x = 3 – 2 = 1

∴ Maximum value of T = x^{2}y = (2)^{2}(1) = 4

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