Problem 4: Linear Programming Using Excel SolverOlympic Bike Ca . is introducing two new lightweight bicycle frames, the Deluxe and the Professional, to be made from special aluminum and steel alloys. The anticipated unit profits are $10 for the Deluxe and $15 for the Professional. A supplier delivers 100 pounds of the aluminum alloy and 80 pounds of the of the steel alloy weekly. The number of pounds of each alloy need per frame is summarized in the following table:\table[[,Aluminum Alloy,Steel Alloy],[Deluse,2,3],[Professional,4,2]]There is an additional constraint in that the number of Deluxe frames produced must be greater than or equal to the number of Professional frames produced. The objective is to maximize total profit subject to the limits on the availability of aluminum and steel and the additional production constraint. Please answer the following questions and submit a working Excel file with the spreadsheet model, as well as the reports requested in part (a) below. Note that the answers for all the questions should be solely based on the Answer Report and Sensitivity Report produced in part (a) below.(a) Solve the linear programming problem using an Excel spreadsheet. Produce the Answer Report and Sensitivity Report from Excel Solvez(b) What is the optimal number of Deluxe frames and Professional frames produced? What is the optimal profit?(c) Suppose the unit profit on the Deluxe frames is increased to $20 per frame. Does the optimal solution change? What is the value of the objective function in this case? Explain your answers.(d) Suppose the unit profit on Deluxe frames is decreased to $6 per frame. Does the optimal solution change? Explain your answer.(e) What is the maximum amount the company pay for an extra pound of aluminum? What is the maximum amount the company should pay for 3 extra pounds of steel? Explain your answers.(f) If the company is forced to produce 1 more Deluxe frame over and above the number of Professional frames produced, how much would total profit change? Explain your answer.