Lecture 11 - Business and Economics Optimization Problems and Asymptotes 11.1 More Economics Applications Price Elasticity of Demand One way economists measure the responsiveness of consumers to a change in the price of a product is with what is called price elasticity of demand. For example, changing the price on vegetables usually strongly affects the demand while changing the price of milk or water doesn t affect that demand that much. Definition 11.1 If p(x is a differentiable demand function, then the price elasticity of demand is given by η = p(x/x dp/dx where η the lowercase Greek letter eta. For a given price, the demand is said to be elastic if η > 1 and the demand is said to be inelastic if η < 1. The demand is unit elasticity if η = 1. The Price Elasticity of Demand η measures the ratio of the percentage change of demand of a product to the percentage change of price of the product. If demand is elastic ( η > 1, the percentage increase in demand is greater than the percentage increase in price, and hence the demand is sensitive to changes in price. Likewise if demand is inelastic( η < 1, the percentage increase in demand is less than the percentage increase in price, and hence the demand is insensitive to changes in price. When elasticity is equal to 1, the percentage changes are roughly equal. For example, rough values of elasticity for some common commodities are: Tomatoes η = 4.60 Automobiles η = 1.35 Housing η = 1.00 Mail η = 0.05 Examples: 1. Let the demand function for a product be modelled by p(x = 21 3 2 x. Find the price elasticity of demand when x = 36 and x = 400.
solution: Here dp = 3 dx 4. So for x = 36: x Thus Here η > 1 so the demand is elastic. For x = 400: Thus Here η < 1 so the demand is inelastic. p(36 = 21 3 2 36 = 12 p (36 = 3 4 36 = 1 8 η = 12/36 1/8 = 8 3 p(400 = 21 3 2 400 = 9 p (400 = 3 4 400 = 3 80 η = 9/400 3/80 = 3 5 2. Suppose the demand function for ice cream bars is given by p(x = 8 2x. For what values of x do ice cream bars have unit elasticity? solution: Mmm, ice cream bars. At any rate, the formula for η is η = p(x/x dp/dx (8 2x/x = 2 = 4 x + 1 We want the values of x for which η = 1. η = 1 4 x + 1 = 1 4 x + 1 = ±1 If 4 x + 1 = 1 then 4 x = 0 which is impossible. If 4 x + 1 = 1 then 4 x = 2 x = 2. Thus x = 2 gives us unit elasticity.
Average Cost Due to the nature of cost functions, the minimum cost usually corresponds to making 0 units. This isn t very useful information, as it is obvious that if we make no units we will spend the least money. More informative would be to find the production level that minimizes the cost per unit, or average cost: Example: C(x = C(x x Suppose that our cost function is C(x = 800 + 0.04x + 0.0002x 2. Find the production level that minimizes the average cost. solution: In this case the averave cost is We find the critical points: C(x = C(x x = 800 x + 0.04 + 0.0002x C (x = 800 x 2 + 0.0002 = 0 x2 = 800 0.0002 = 4,000,000 Since we are again only considering x in (0,, we have one critical point, x = 2000. Now, C (x = 1600 x 3 C (2000 = 1600 2000 > 0 3 so x = 2000 is a local minimum, and that together with the fact that it is the only critical point in the domain (0, gives us that it is an absolute minimum. The minumum average cost is C(2000 = 800 + 0.04 + 0.0002(2000 = $0.84 2000 Example: (Old Midterm Question A pumpkin patch can sell 100 pumpkins per week at a price of $3 each. The pumpkin farmer estimates that for each $0.10 reduction in price, she can sell 5 more pumpkins per week. (a Assuming the demand function is linear, find the demand function and the total revenue as a function of the number of pumpkins sold. (b How many pumpkins does she sell when revenue is maximized? (c What price per pumpkin maximizes the revenue?
(d Explain why your answer is an absolute maximum. solution: (a We are assuming the demand function p(x = mx + b. To find m we use the formula m = p x = 0.10 5 So p(x = 1 x + b. To find b we plug in a point 50 = 1 50 3 = 1 50 (100 + b b = 5 So p(x = 1 1 x + 5, and thus R(x = xp(x = 50 50 x2 + 5x. (b We want to maximize R(x. The domain of R(x is [0,. We find the critical points of R: R (x = 1 25 x + 5 = 0 x = 125 So x = 125 is our only critical point. R (125 = 1 < 0, so x = 125 is a local 25 maximum. This together with the fact that it is the only critical point in [0, implies that it is an absolute maximum. (c We plug in x = 125 p(25 = 1 50 (125 + 5 = 5 2 + 5 = 2.5 So setting the price to be $2.50 maximizes revenue. (d It is explained in (b above. The point x = 125 is a local maximum, and this together with the fact that it is the only critical point in [0, implies that it is an absolute maximum.
11.2 Asymptotes The function f(x = 3 x 2 has a graph that looks like this Here we say that x = 2 is a vertical asymptote for the graph of f. Notice that f(x = x 2 + This leads to the following definition. f(x = x 2 Definition 11.2 If f(x approaches or as x approaches some fixed c from the right or left, then the line x = c is said to be a vertical asymptote of the graph of f. If f is a rational function, ie f(x = g(x h(x then c is a vertical asymptote for f if h(x = 0 and g(x 0. Examples: 1. Suppose that f is defined as f(x = x + 4 x 2 4x = x + 4 x(x 4 Then the denominator is 0 when x = 0 or x = 4. Neither of these make the numerator 0, so both x = 0 and x = 4 are vertical asymptotes. To find out the behaviour of f
near the asymptotes we compute the one-sided its: f(x x 0 = x + 4 x 0 x(x 4 = + f(x x 0 + = = f(x x 4 = x + 4 x 4 x(x 4 = f(x x 4 + = = + x + 4 x 0 + x(x 4 x + 4 x 4 + x(x 4 ( + ( ( ( + (+( ( + (+( ( + (+(+ We summarize this with the following graph 2. Suppose that f is defined as f(x = x2 + 2x 8 x 2 4 = (x 2(x + 4 (x 2(x + 2 In this case we have a vertical asymptote at x = 2 but not at x = 2. Note that f is still not defined at the point x = 2 even though the (x 2 term cancels out. When
x 2, we can cancel (x 2(x + 4 f(x = (x 2(x + 2 = x + 4 x + 2 x 2 = x + 2 + 2 x + 2 x 2 = 1 + 2 x + 2 x 2 So we see that everywhere except x = 2, f looks like 1 + 2. At x = 2, f has a hole. x+2 Notice that the last function appears to approach a value as x goes to. This leads to the following definition. Definition 11.3 If f is a function and L is a real number, the statements f(x = L or f(x = L x denote its at. In either of these cases, the line y = L is called a horizontal asymptote for f. Examples: 1. For any r > 0, 2. In the last example with the graph above, f(x = x2 + 2x 8 x 2 4 1 x r = 0 = 1 + 2 x + 2 for x 2
2 If we take the it as x approaches, the term gets smaller and smaller, and so x+2 ( 1 + 2 = 1 x + 2 So the function has a horizontal asymptote at y = 1. A common situation is finding horizontal asymptotes for rational functions p(x. There are q(x a few simple rules that will let you easily see what asymptotes such functions have. 3. Find the horizontal asymptotes of f(x = 2x+1 4x 2 +5. solution: To evaluate this it, we multiply the top and bottom by the reciprocal of the largest power of x involved (in this case, that will be 1 x 2 : ( ( 2x + 1 2x + 1 1 4x 2 + 5 = x 2 4x 2 1 + 5 2 = + 1 x x 2 4 + 5 x 2 = 0 + 0 4 + 0 = 0 So f(x has a horizontal asymptote at y = 0. 4. Find the horizontal asymptotes of g(x = 2x2 +1 4x 2 +5 solution: We can use the same trick as before: 2x 2 + 1 4x 2 + 5 = ( 2x 2 + 1 4x 2 + 5 = 2 + 1 x 2 4 + 5 x 2 = 2 + 0 4 + 0 = 1 2 So g(x has a horizontal asymptote at y = 1 2. x 2 ( 1 x 2 1 x 2 The last couple examples can be summarized as follows: suppose f(x = p(x, where p and q(x q are both polynomials. Then If the degree of p is less than the degree of q, then f has a horizontal asymptote at y = 0.
If the degree of p is equal to the degree of q, then f has a horizontal asymptote at y = L where L is the ratio of the leading coefficient of p to the leading coefficient of q. If the degree of p is greater than the degree of q, then the function has no horizontal asymptotes. Example: (Real life example of asymptotic behaviour A small business invests $5000 in a new product. In addition to the initial investment, the product costs $0.50 per unit to produce. Find the average cost per unit a if 1000 units are produced, b if 10000 units are produced, c as the number of units produced goes to infinity. solution: We can just read the formula off and evaluate at 1000 and 10000 to get a and b To get c we evaluate C(x as x : C(x = 5000 + x 2 C(x = 5000 x + 1 2 C(1000 = 5 + 1 2 = $5.50 C(10000 = 1 2 + 1 2 = $1 C(x = 0 + 1 2 = $0.50 This makes sense; as you make more and more units the cost due to the initial investment becomes less and less significant.