Problem 1 A common problem arising in engineering applications is to maximize or minimize a function of several variables. Although in many cases maxima and minima can be found analytically by setting partial derivatives to zero, as discussed in class, the resulting equations for the independent variables are, in general, difcult to solve. In cases Where one or more coefficients vary and solutions need to be recomputed frequently, it is typically more efcient to use a direct optimization method. In this problem you will explore one of the simplest gradient- based methods, usually known as the method of steepest descent. Suppose the problem is to minimize a function of two variables 20%)?) Without constraints. Recall that the gradient of a inction points in the direction of its maximum change. Therefore, if one were to start with an initial guess (xoayu) and move in the direction opposite to the direction of the gradient V2 2 az/ax i +6z/6yj , the value of the flmction z(x, y) at the new point (x1, yl) should be smaller than the value of z(x, y) at 060,320). In fact, the direction opposite to the direction of the gradient is the best direction to move in to locate the minimum. Thus, at every iteration n, the coordinates can be updated according to the following rule: [62] xn+l : xn _ t 6x n y _, {5} n+1 n 6)} n Where t is an arbitrary parameter that determines the step size. a) Consider the following function: z(x, y) = (x - Ax - y)e where A is a coefficient that can vary between 1 and 2. Assuming A = 1.5 make two separate plots of z(x, y) and its level curves over the interval 0