Please Solve Problem 5 Only, use Solver constrained optimmization. The function

Question

Please Solve Problem 5 Only, use Solver constrained optimmization.

The function shown below has a minimum at [0, 0]. From an initial estimate of [1, 1], perform one steepest ascent step (on negative f{x, y}) for your estimate of the minimum f (remember that you must negate your final result). Use bisection method to find the optimum step size h in the gradient direction (where the function dF/dh = 0) with a stopping criterion of 1% on h. f(x, y) = x^4 + (x + y)^2 + (e^y - 1)^2 Using the function and initial estimate of Problem 3 (above), perform 4 Newton iterations toward the minimum. Note that you do not need to negate the function, because the Newton method looks only at derivatives (which are not sign sensitive). Using the function and initial estimate of Problem 3 (above) to directly use Solver to minimize the function f(x, y).

Explanation / Answer

5.

f(x,y) = x^4 + (x+y)^2 + (e^y-1)^2

first we'll have to find the critical point(s):

df/dx = 4x^3 + 2(x+y)

df/dy = 2(x+y) + 2(e^y-1)

now plug df/dx = 0 an df/dy =o and solve for x and y

=> 4x^3 + 2(x+y) = 0

and 2(x+y) + 2(e^y-1) = 0

solving the above equations simultaneously

we get the critical point as (0,0)

now in order to nind the kind of critical point (0,0) is that is whether it a

maximax or minima or a saddle point we'll have to find the double derivative of the function f(x,y)

d^2f/dx^2 = 12x^2 + 2

d^2f/dxdy = 2

d^2f/dy^2 = 2 + 2e^y

d^2f/dx^2(0,0) = 2

d^2f/dxdy(0,0) = 2

d^2f/dy^2(0,0) = 2 + 2e^y = 3

now D = d^2f/dx^2(0,0) * d^2f/dy^2(0,0) - d^2f/dxdy(0,0)

D = 2*3 - 2 = 4

=> D >0

and d^2f/dx^2(0,0) > 0

=> the critacal point (0,0) is a point of relatie minimum value.

that is at (0,0) we'll get a relative minimum function value

=> the minimum value of the function f(x,y) is

f(x,y) = x^4 + (x+y)^2 + (e^y-1)^2

f(0,0) = 0

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