The problem is to maximize the function f(x; alpha ) = ax^2-(20 alpha +2)x, subj
ID: 1191472 • Letter: T
Question
The problem is to maximize the function f(x; alpha ) = ax^2-(20 alpha +2)x, subject to the constraints where alpha is an exogenous constant. For parts (d) and (e), you may want to check your answers by confirming that they reproduce your answers to (a) and (b). Assume alpha =-1. State the first-order condition, and find any points that satisfy it, within the constraints. Does f(x; 1) satisfy the global second-order condition? What value of x solves the problem? (Justify your answer carefully.) What is the maximized value of f(x;-1)? Repeat part (a), this time assuming alpha = 1. For which values of a is there a point that satisfies the FOC, within the constraints? Calculate a function x(a), which shows the solution to the problem, given any real number alpha . (It may be helpful to consider the range of values of a identified by part (c), separately from other values of alpha .) Draw a graph of your function, showing the exact coordinates of at least three points that have different values for x( alpha ). Calculate a function M(alpha), which shows the maximized value of f that can be achieved, given any real number alpha. Draw a graph of your function, showing the exact coordinates of M(alpha) for the same values of a that you used in part (d).Explanation / Answer
With a= -1, f(x,1) = x2-18x
Lagrangian equation,
f(x,1) = x2-18x
First order conditions, 2x - 18 =0
=> x=9
SOC: f''(9) = 2 >0 , x=9 gives local minimum value
f(0) =0
f(10) = -80
f(9) = -82
x=9 is global minima and x=0 is global maxima.
b) a=1,
f(x,1) = x2 -22x
df/dx = 2x -22 = 0 => x =11
It is not within the constraint.
f(0) = -22
f(10) = -2
Global maxima is x=10 and global minima is x=0.
f''(10) = 2>0
It does not satisfy the second order condition.
c) For a = -1
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