Assuming a linear demand curve, find the price of a hamburger that will maximize

Question

Assuming a linear demand curve, find the price of a hamburger that will maximize the nightly hamburger revenue. If the concessionaire has fixed costs of $1000 per night and the variable cost is $.60 per hamburger, find the price of a hamburger that will maximize the nightly hamburger profit. Demand and Revenue The average ticket price for a concert at the opera house was $50. The average attendance was 4000. When the ticket price was raised to $52, attendance declined to an average of 3800 persons per performance. What should the ticket price be to maximize the revenue for the opera house? (Assume a linear demand curve.) Demand and Revenue An artist is planning to sell signed prints of her latest work. If 50 prints are offered for sale, she can 50 prints, she must lower the price of all the prints by $5 for each print in excess of the 50. How many prints should the artist make to maximize her revenue? Demand and Revenue A swimming club offers memberships at the rate of $200, provided that a minimum of 100 people join. For each member in excess of 100, the membership fee will be reduced $1 per person (for each member). At most, 160 memberships will be sold. How many memberships should the club try to sell to maximize its revenue? Profile In the planning of a sidewalk cafe, it is estimated that for 12 tables, the daily profit will be $10 per table. Because of overcrowding, for each additional table the profit per table (for every table in the cafe) will be reduced by $.50. How many tables should be provided to maximize the profit from the cafe? Demand and Revenue A certain toll road averages 36,000 cars per day when charging $1 per car. A survey concludes that increasing the toll will result in 300 fewer cars for each cent of increase. What toll should be charged to maximize the revenue?

Explanation / Answer

We define x to be the number of painting the artist paints and y the price of each painting. We want to maximize the revenue. The revenue is the cost per painting multiplied by the number of paintings sold, so our objective equation is: R = x · y So, the idea is that if the price lowers by $5 for every painting we make after 50, we have a nice linear relationship with slope 5, we also have the point (50, 400). Using the point-slope formula, we have y 400 = 5(x 50) y 400 = 5x + 250 y = 5x + 650 Which gives us our constraint equation. Now, we can plug this into the objective equation to get: R(x) = x(5x + 650) = 5x 2 + 650x Now, we want to take the derivative of this and set it equal to zero. R 0 (x) = 10x + 650 0 = 10x + 650 10x = 650 x = 65

Now, to make sure this is indeed a maximum, we take the second derivative of the revenue function and evaluate it at x = 65. R 00(x) = 10 < 0 Thus, the artist should make 65 prints to maximize her revenue.

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