What is the maximum value of x if x = y^2 + 8y + 16
1 Answer | Add Yours
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.
Join to answer this question
Join a community of thousands of dedicated teachers and students.Join eNotes