Use the revenue function from Example 3 and the cost function from Matched Problem 3:

Both have domain 1 ≤ x ≤ 15.

(A) Sketch the graphs of both functions in the same coordinate system.

(B) Break-even points are the production levels at which R(x) = C(x). Find the break-even points algebraically to the nearest thousand cameras.

(C) Plot both functions simultaneously in the same viewing window.

(D) Use intersect to find the break-even points graphically to the nearest thousand cameras.

(E) Recall that a loss occurs if R(x) C(x). For what values of x (to the nearest thousand cameras) will a loss occur? A profit?

Data from Example 3

Recall that the financial department in the company that produces a digital camera arrived at the following price–demand function and the corresponding revenue function:

where p(x) is the wholesale price per camera at which x million cameras can be sold and R(x) is the corresponding revenue (in millions of dollars). Both functions have domain 1 ≤ x ≤ 15.

Find the value of x to the nearest thousand cameras that will generate the maximum revenue. What is the maximum revenue to the nearest thousand dollars? Solve the problem algebraically by completing the square.

What is the wholesale price per camera (to the nearest dollar) that generates the maximum revenue?

Graph the revenue function using an appropriate viewing window.

Find the value of x to the nearest thousand cameras that will generate the maximum revenue. What is the maximum revenue to the nearest thousand dollars? Solve the problem graphically using the maximum command.

Data from Matched Problem 3

The financial department in Example 3, using statistical and analytical techniques arrived at the cost function

where C(x) is the cost (in millions of dollars) for manufacturing and selling x million cameras.

Transcribed Image Text:

R(x) = x(94.8 - 5x) C(x) = 156 + 19.7x Revenue function Cost function