The given cost function of the firm is,C = q3 - 36q2 + 490q + 1000The market price of the product is given as, p = $250We know that the revenue function R is given by,R = pqWhere q is the quantity produced.The firm's profit function is given by the equation,P = R - CThus, we need to first calculate the revenue function and then substitute the values of R and C in the profit function to find out the profit function and other parameters. The revenue function is,R = pqR = 250qThe firm's profit function is,P = R - CP = 250q - (q3 - 36q2 + 490q + 1000)P = -q3 + 36q2 + 240q - 1000We need to maximize the profit function P, so we differentiate it with respect to q and equate it to zero.dP/dq = -3q2 + 72q + 240= 0On solving the above quadratic equation, we get,q = 10 or q = 8We now find the second derivative of the profit function,P'' = -6q + 72P''(10) = -6(10) + 72 = 12Therefore, q = 10 is a point of maximum profit. To check whether it is a maximum point or not, we find the value of P''(10) which is positive. Hence, q = 10 is the quantity that maximizes the profit. Substituting the value of q in the profit function, we get, P = 16000The firm's profit is $16,000. As the profit is positive, the firm should operate rather than shutting down.