Maximizing Profit A company is interested (naturally) in determining the number
ID: 2860866 • Letter: M
Question
Maximizing Profit A company is interested (naturally) in determining the number of items to manufacture in order to maximize their profits. To make this determination, v/e need to find the difference between the company's revenues and costs. In basic economics, a typical cost function is given by a cubic polynomial, i.e., a function of the form C(x) = ax^3 + bx^2 + cx + d, where x represents the quantity of items a company manufactures (and sells). The number d represents the fixed costs, or the cost incurred even if no items are produced, and the terms ax^3 + bx^2 + cx are the variable costs, or the costs that change depending on how many items are produced. These variable costs take into account factors including materials and labor (along with the possibility, say, of gening a better deal the more materials you purchase, or diminishing productivity the more labor is required). Revenue depends on the amount of items that are sold, or the demand for an item. This demand depends on the price of the item, e.g., a higher price induces lower demand. Let's say the demand function for an item is given by x = 300 - 1/6 p, v/here x is the number of items demanded at a price p. Find the revenue function, i.e., a function that gives the amount of income from selling x items. Suppose this company has a cost function of C(x) = 0.004x^3 - 1.6x^2 + 500 x + 16,000 for producing x many items. Given your revenue function from part (a), find the profit function, i.e., the amount of money you have after expenses, and maximize this function.Explanation / Answer
(a) Revenue = x* p
Now we have x = 300-(1/6)p => x-300 = p/6 => 6(x-300) = p
Hence R(x) = x*6(x-300) = 6x(x-300)
(b) Profit function = R(x) - C(x) = 6x2-1800x-0.004x3+1.6x2-500x-16000 = -0.004x3+7.6x2-2300x-16000
Now P'(x) = -0.012x2+15.2x-2300 = 0
x = 175.68 and 1091
Now in (0,175.68) consider a test point x= 1, then P'(x) = -0.012(1)+15.2(1)-2300 = -2284.812<0
In (175.68, 1091) consider a test point x= 1000, the P'(1000) = -0.012(1000)2+15.2(1000)-2300 = -12000+15200-2300 = 15200-14300 >0
Also consider (1091, infinity) let x = 2000, then P'(2000) = -48000+30400-2300 <0
As sign change of P'(x) is from >0 to <0 for 1091
Hence x=1091 gives maximum value of P(x) and P(1091) = -5194386.284+9046135.6-2509300-16000
=> Max P = 1326449.316