The force F (in newtons) of the hydraulic cylinder in a press is proportional to the square of sec x where x is the distance (in meters) that the cylinder is extended in its cycle. The domain of F is [0, pi/3], and F(0) = 500.A) find F as a function of x.F(x)=___________B) find the average force exerted by the press over the interval [0, pi/3] (round your answer to 1 decimal place)F= _________? N

Answer:Part A) F(x) =500 sec(x)Part B) Average force exerted by the press = 750 NStep-by-step explanation:Given:F(x) ∝ sec(x)Range of x = [0,Pi/3] or x = [0 , 60°]F(0) = 500 N i.e. F is 500 N at x = 0.As we now that:Force is proportional to sec(x) therefore we may write an equation by introducing Proportionality Constant (k) as under:F(x) = k * sec(x) - Say it Equation 1Then using given information of F=500 N at x = 0 we have:500 = k * sec(0)As sec(0) = 1, therefore we get:k = 500By putting value of k in equation 1 we have:F(x) = 500 * sec(x)Now by putting max value of x from the given range that is Pi/3 in the above equation we get:F(Pi/3) = 500 * sec(Pi/3)As Pi = 180 there we simplify the above equation as:F(60°) = 500 * sec(60°)F(60°) = 500 * 2 ; By putting sec(60°) = 2F(60°) = 1000 NNow the avg. of force exerted by the hydraulic press is given by:Avg. Force = Minimum Force + [(Maximum Force - Minimum Force) / 2] - Say it equation 2Considering sec(x) is minimum at x = 0 and maximum at x = 60° within the given range [0 , 60°] therefore we have:The minimum force being put at x = 0 i.e. F = 500 N and max. force at x = 60° which is F = 1000 N.Finally we get the average force using equation 2 as under:Average force = 500 +  [(1000 - 500) / 2]Average Force = 500 + 250Average Force = 750 N