1) Find the values of x1 and x2 that maximize f(x1, x2) = 5 + In(x1) + In(x2) such that 2x,? + x,? <3, x1 2 0, and x2 2 0.2) If we add more constraints to the above problem, how will that affect the optimal value of theobjective? That is, will the maximum value of the objective become larger, smaller, or unchanged ifextra constraints are added to the above problem?

Given: The objective function, fx1,x2=5+lnx1+lnx2 The primary constraint 2x12+x22-3≤0 and non-negativity constraints x1≥0 and x2≥0. (1.) To find: The values of x1 and x2 that maximize the given function with respect to the given constraints. (2.) To determine: How the maximum value of the objective function varies if more constraints are added to the given problem.(1.) Given function: fx1,x2=5+lnx1+lnx2 Given constraints: 2x12+x22-3≤0, x1≥0 and x2≥0 Now, maximizing fx,y is equivalent to minimizing -fx,y Clearly, -fx1,x2=-5-lnx1-lnx2 Then, the equivalent minimization problem to the given maximization problem is given as, Minimize -fx1,x2=-5-lnx1-lnx2 subject to the constraints 2x12+x22-3≤0 x1≥0, x2≥0 We can convert any inequality a≤0 to an inequality by adding a positive quantity to the left-hand side as follows: a+y2=0 for some y Then, the equivalent minimization problem with equality constraint is, Minimize -fx1,x2=-5-lnx1-lnx2 subject to the constraint Gx1,x2,y=gx1,x2+y2=0, where gx1,x2=2x12+x22-3 x1≥0, x2≥0 We will solve this minimization problem using the process of Lagrange multiplier. The Lagrange function for this problem can be written as, Lx1,x2,y=-fx1,x2+λGx1,x2,y, where λ is the Lagrange multiplier. Put gx1,x2=2x12+x22-3 in Gx1,x2,y=gx1,x2+y2 to get, Gx1,x2,y=2x12+x22-3+y2 Put -fx1,x2=-5-lnx1-lnx2 and Gx1,x2,y=2x12+x22-3+y2 in Lx1,x2,y=-fx1,x2+λGx1,x2,y to get, Lx1,x2,y=-5-lnx1-lnx2+λ2x12+x22-3+y2 This is the required Lagrange function.The Lagrange function has been obtained as, Lx1,x2,y=-5-lnx1-lnx2+λ2x12+x22-3+y2 Now, evaluating the partial derivatives of the Lagrange function as follows: ∂L∂x1=∂∂x1-5-lnx1-lnx2+λ2x12+x22-3+y2=-∂∂x1lnx1+λ∂∂x12x12+x22-3+y2=-1x1+λ∂∂x12x12=-1x1+4λx1 So, ∂L∂x1=-1x1+4λx1 Similarly, ∂L∂x2=∂∂x2-5-lnx1-lnx2+λ2x12+x22-3+y2=-∂∂x2lnx2+λ∂∂x22x12+x22-3+y2=-1x2+λ∂∂x2x22=-1x2+2λx2 So, ∂L∂x2=-1x2+2λx2 Similarly, ∂L∂λ=∂∂λ-5-lnx1-lnx2+λ2x12+x22-3+y2=∂∂λλ2x12+x22-3+y2=2x12+x22-3+y2 So, ∂L∂λ=2x12+x22-3+y2 Finally, ∂L∂y=∂∂y-5-lnx1-lnx2+λ2x12+x22-3+y2=λ∂∂y2x12+x22-3+y2=λ∂∂yy2=2λy So, ∂L∂y=2λy According to the optimality conditions, we have, ∂L∂x1=∂L∂x2=∂L∂λ=∂L∂y=0 Put ∂L∂x1=0 in ∂L∂x1=-1x1+4λx1 to get, -1x1+4λx1=0⇒4λx1=1x1⇒λ=14x12 So, λ=14x12 Put ∂L∂x2=0 in ∂L∂x2=-1x2+2λx2 to get, -1x2+2λx2=0⇒2λx2=1x2⇒λ=12x22 So, λ=12x22 Put ∂L∂λ=0 in ∂L∂λ=2x12+x22-3+y2 to get, 2x12+x22-3+y2=0 Put ∂L∂y=0 in ∂L∂y=2λy to get, 2λy=0λy=0 This implies that either λ=0 or y=0. However, if we consider λ=0, then the Lagrange function Lx1,x2,y=-fx1,x2+λGx1,x2,y reduces to Lx1,x2,y=-fx1,x2 and the optimality of the Lagrange function denotes the global optimality of the objective function without taking into account the given constraint. Thus, if λ=0, then the given constraint becomes inactive in the minimization problem. So, to keep the constraint active, we must consider λ≠0 and thus y=0. Now, we have obtained the following equations from the optimality conditions: λ=14x12 λ=12x22 y=0 2x12+x22-3+y2=0 Equating the values of λ from λ=14x12 and λ=12x22, we have, 14x12=12x22⇒12x12=1x22⇒x22=2x12⇒x2=±2x1 So, x2=±2x1 Put y=0 in 2x12+x22-3+y2=0 to get, 2x12+x22=3 Put x2=±2x1 in 2x12+x22=3 to get, 2x12+2x12=3⇒4x12=3⇒x12=34⇒x1=±32 However, one of the non-negativity constraint is given as x1≥0. Then, x1=32 Put x1=32 in x2=±2x1 to get, x2=±232=±32 However, one of the non-negativity constraint is given as x2≥0. Then, x2=32 So, x1=32 and x2=32 are the values that maximize the given function with respect to the given constraints.(2.) Let us assume that z is the optimal value of the objective function obtained by solving the given problem. Then z is the maximum value of fx1,x2=5+lnx1+lnx2 subject to the given constraints. Now, if we add one more constraint to the problem, there are two possible cases. The first case is when our original solution satisfies the new constraint. In this case, the original solution is also the solution of the new problem with the added constraint. The second case is when the original solution does not satisfy the new constraint. In this case, either the new problem (with the added constraint) is not feasible, or there is a new solution. Now, the new solution must satisfy both the new constraint and the original constraint. However, the maximum value of fx1,x2=5+lnx1+lnx2 subject to the original constraint is z. Since the new solution also satisfies the original constraint, it cannot give a maximum value for fx1,x2=5+lnx1+lnx2 greater than z. Thus, the new solution must give a maximum value of fx1,x2=5+lnx1+lnx2 lesser than z. Continuing to add more constraints, the maximum value of fx1,x2=5+lnx1+lnx2 thus either remain constant or decreases. Thus, if we add more constraints to the given problem, then the maximum value of the given function either remain unchanged or become smaller.Answer: (1.) It has been determined that x1=32 and x2=32 are the values that maximize the given function with respect to the given constraints. (2.) It has been determined that if we add more constraints to the given problem, then the maximum value of the given function either remain unchanged or become smaller.