Is it possible to solve the value of lambda that maximizes the following equation analytically?: 1 − e − 30 λ 30 λ − e − 30 λ So then the derivative is equal to: e − 30 λ λ − 30 900 λ 2 + 30 e − 30 λ 900 λ 2 + 30 e − 30 λ = 0Is it possible to get an analytical solution for lambda or does it need to be solved numerically?

Question

Is it possible to solve the value of lambda that maximizes the following equation analytically?:            1      −              e                  −          30          λ                            30      λ        −      e          −      30      λ      So then the derivative is equal to:            e              −        30        λ              λ    −      30          900              λ        2              +            30              e                  −          30          λ                            900              λ        2              +  30      e          −      30      λ        =  0Is it possible to get an analytical solution for lambda or does it need to be solved numerically?

humusen6p · Accepted Answer

Calling   y  =  30  λ we have  f  (  y  )  =      1    y    −      (          1      y        +    1    )        e          −      y      now      f    ′    (  y  )  =                    e                  −          y                            (                  y          2                +        y        −                  e          y                +        1        )                    y      2        =  0so making approximations  1  +  y  +      y    2    −      e    y    ≈            y      2        2    −            y      3              3      !        −            y      4              4      !        =  0  →      1    2    =      y    6    +            y      2        24  and solving we obtain   y  =  {  −  6  ,  2  } so with a sufficient approximation the maximum is located at   y  =  30  λ  =  2  →  λ  =      1    15

Is it possible to solve the value of lambda that maximizes the following equation analytically?:

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