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What is the maximum value of x if x = y^2 + 8y + 16

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What is the maximum value of x if x = y^2 + 8y + 16

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The maximum value of x has to be determined if x = y^2 + 8y + 16.

The function x = y^2 + 8y + 16 has a maximum value at the value of y = c that is the solution of `dx/dy = 0` and `(d^2x)/dy^2` should be negative for y = c.

`dx/dy = 2y + 8`

`(d^2x)/dy^2 = 2` which is positive for all y.

This indicates that it is not possible to determine a positive value for x = y^2 + 8y + 16 as x can tend to infinity as x tends to infinity.

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