Multiple Choice
A poster is set to have a total area of 1150 cm2, with 2-cm margins on the sides and the top, and a 3-cm margin at the bottom. What dimensions will maximize the printed area?
6. Graphical Applications of Derivatives
Applied Optimization
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Your café sells lattes for $4 each to 100 customers per day. For every $1 increase in price, you would lose 20 customers. Find the price that maximizes revenue. Hint: The # of items sold is based on the number of customers.
A
$4.00
B
$4.50
C
$5.00
D
$5.50
Verified step by step guidance1
Step 1: Define the variables. Let the price of a latte be denoted as p (in dollars). The number of customers is a function of the price, and it decreases by 20 for every $1 increase in price. If the base price is $4 with 100 customers, the number of customers can be expressed as 100 - 20(p - 4).
Step 2: Write the revenue function. Revenue is the product of the price per latte and the number of lattes sold. Using the expression for the number of customers, the revenue function R(p) is given by R(p) = p * (100 - 20(p - 4)).
Step 3: Simplify the revenue function. Expand the terms in R(p) to express it as a quadratic function. This will help in identifying the price that maximizes revenue. Simplify R(p) = p * (100 - 20p + 80) to R(p) = -20p^2 + 180p.
Step 4: Find the critical points. To maximize revenue, take the derivative of R(p) with respect to p, denoted as R'(p), and set it equal to zero. Compute R'(p) = d/dp(-20p^2 + 180p) = -40p + 180. Solve the equation -40p + 180 = 0 to find the critical point.
Step 5: Verify the maximum. Use the second derivative test to confirm that the critical point corresponds to a maximum. Compute the second derivative R''(p) = d^2/dp^2(-20p^2 + 180p) = -40. Since R''(p) is negative, the function has a maximum at the critical point. Substitute the critical point back into the revenue function to find the maximum revenue and corresponding price.
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