I am not sure how to solve these two, especially using the simple method? Can someone help! ### Problem 6: HARMATHAP12 4.5.015**Objective:**Use the simplex method to find the optimal solution. Assume all variables are nonnegative.**Minimization Problem:**Minimize \( f = 3x + 2y \)**Subject to the Constraints:**- \(-x + y \leq 15\)- \(x + y \geq 35\)- \(x + y \leq 45\)**Variables to Determine:**- \(x = \) [input area]- \(y = \) [input area]- \(f = \) [input area]### Problem 7: HARMATHAP12 4.5.020**Objective:**Use the simplex method or Excel. Assume that all variables are nonnegative.**Maximization Problem:**Maximize \( f = -x + 3y + 4z \)**Subject to the Constraints:**- \(x + y + z \leq 40\)- \(-x + y + z \geq 30\)- \(x + y - z \geq 20\)**Variables to Determine:**- \(x = \) [input area]- \(y = \) [input area]- \(z = \) [input area]- \(f = \) [input area]

Note: Since you have asked multiple questions, we will solve the first question for you. If you want any specific question to be solved then please specify the question number or post only that question.” Introduction: The Simplex method may be a method for manually solving applied mathematics models employing slack variables, tableaus, and pivot variables to spot the most effective solution to an optimization issue. Simplex tableau is employed to test optimality and conduct row operations on the applied math model.The surplus variable represents the number of resource usage above the minimum required and is added to greater-than or equal-to constraints so as to urge equality constraint.Given: It is given that the linear programming problem is, min f=3x+2ysubject to-x+y≤15x+y≥35x+y≤45 We have to solve this problem by the simplex method. First, add slack, surplus, and artificial variables to the L.P.P. min f=3x+2y+0.S1+0.S2+0.S3+MA1subject to-x+y+S1=15x+y-S2+A1=35x+y+S3=45Now, make the first iteration table. Iteration-1 Cj 3 2 0 0 0 M B CB XB x1 x2 S1 S2 S3 A1 MinRatioXBx2 S1 0 15 -1 (1) 1 0 0 0 151=15→ A1 M 35 1 1 0 -1 0 1 351=35 S3 0 45 1 1 0 0 1 0 451=45 Z=35M Zj M M 0 -M 0 M Zj-Cj M-3 M-2↑ 0 -M 0 0 Here pivot element is (1) and the entering element is x2.Now, make the second iteration table. Iteration-2 Cj 3 2 0 0 0 M B CB XB x1 x2 S1 S2 S3 A1 MinRatioXBx1 x2 2 15 -1 1 1 0 0 0 --- A1 M 20 (2) 0 -1 -1 0 1 202=10→ S3 0 30 2 0 -1 0 1 0 302=15 Z=20M+30 Zj 2M-2 2 -M+2 -M 0 M Zj-Cj 2M-5↑ 0 -M+2 -M 0 0 Here pivot element is (2) and the entering element is x1.Now make the third iteration table. Iteration-3 Cj 3 2 0 0 0 B CB XB x1 x2 S1 S2 S3 MinRatio x2 2 25 0 1 0.5 -0.5 0 x1 3 10 1 0 -0.5 -0.5 0 S3 0 10 0 0 0 1 1 Z=80 Zj 3 2 -0.5 -2.5 0 Zj-Cj 0 0 -0.5 -2.5 0 Here every zj-Cj≤0, thus the optimal solution has come and value of x1=10 and x2=25. Substitute x1=10 and x2=25 in f=3x+2y. f=3(10)+2(25)=80Hence the required solution is as follows: x1=10x2=25f=80