ma132.1 - Math 132 Lesson 1 Declining Price Profits and...

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Math 132 Lesson 1: Declining Price, Profits and Graphing Application A clothing firm can manufacture suits at 20 dollars per suit. The initial or startup cost of the equipment is 10,000 dollars. The revenue was initially 80 dollars per suit, the selling price. But as more suits were sold, the demand dropped. The retailer noticed a trend in the declining price: for every additional suit sold the price had to be dropped by about 5 cents (.05 dollars). The objective is to find the break-even number of sales where revenue is equal to cost. If the sales are too small, the initial cost will make the total cost larger than the revenue. In the other extreme, if the sales are too large, the price per suit will be very small and the revenue will be too small to realize a profit. Also we would like to know the number of sales where the profit will be a maximum. Math Model The profitis defined to be the revenueminus the total costs: [profit] = [revenue] - [total cost] The revenueis the price per suit times the number of suits sold: [revenue] = [price per suit]*[number sold] The total cost for producing x suits is [total cost] = [production cost of suits] + [initial cost] [production cost of suits] = [cost to make each suit]*[number manufactured] We are ready to translate our word equations to symbolic equations. We have to define our variables: x = number of suits sold C(x) = total cost R(x) = revenue P(x) = profit S(x) = price per suit So, translating our word equations into symbolic equations, we get: [profit] = [revenue] - [total cost] P(x) = R(x) - C(x) [revenue] = [price per suit]*[number sold] R(x) = S(x)*x [total cost] = [production cost of suits] + [initial cost] [production cost of suits] = [cost to make each suit]*[number manufactured] C(x) = 20x + 10,000 We still need S(x). The price per suit was initially 80 but declined by .05 dollars for each additional In applying math to the real world, we have to translate concepts to symbols and back again. It can be very helpful to take an intermediate step and construct word equationslike these. In applying math to the real world, a very important part is defining your variables. If your definitions aren't precise, you'll get confused. We couldn't have used P for profit and P for price, because then we would have one symbol meaning two different things.

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