the annual total revenue for a product is given by ( r(x) = 63,000x - 7x^{2} ) dollars, where ( x ) is the…

the annual total revenue for a product is given by ( r(x) = 63,000x - 7x^{2} ) dollars, where ( x ) is the number of units sold. to maximize revenue, how many units must be sold? what is the maximum possible annual revenue?\nto maximize revenue, ( square ) units must be sold.\n(simplify your answer.)

the annual total revenue for a product is given by ( r(x) = 63,000x - 7x^{2} ) dollars, where ( x ) is the number of units sold. to maximize revenue, how many units must be sold? what is the maximum possible annual revenue?\nto maximize revenue, ( square ) units must be sold.\n(simplify your answer.)

Answer

Explanation:

Step1: Find the derivative of the revenue function

The revenue function is ( R(x) = 63000x-7x^{2}). Using the power rule ((x^n)^\prime=nx^{n - 1}), the derivative (R^\prime(x)) is: (R^\prime(x)=\frac{d}{dx}(63000x)-\frac{d}{dx}(7x^{2})) (R^\prime(x)=63000-14x)

Step2: Set the derivative equal to zero and solve for (x)

To find the critical points, set (R^\prime(x) = 0). (63000-14x=0) Add (14x) to both sides: (14x = 63000) Divide both sides by (14): (x=\frac{63000}{14}=4500)

Step3: Check the second - derivative

The second - derivative (R^{\prime\prime}(x)=\frac{d}{dx}(63000 - 14x)=-14) Since (R^{\prime\prime}(x)=-14<0), the function (R(x)) has a maximum at (x = 4500)

Answer:

(4500)