This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Chapter 6 Optimization In this chapter, we collect a variety of problems in which the ideas developed in earlier material are put to use. In particular, we will use calculus to find local (and global) maxima, and minima so as to get the best (optimal) values of some desirable quantity. Setting up these problems, from first verbal description, to clear cut mathematical formulation is the main challenge we will face. Often, we will use geometric ideas to express relationships between variables leading to our solution. 6.1 Biological optimization 6.1.1 Density dependent growth in a population Biologists often notice that the growth rate of a population depends not only on the size of the population, but also on how crowded it is. When individuals have to compete for resources, nesting sites, mates, or food, they cannot reproduce as quickly, leading to a decline in the rate of growth of the population. The rule that governs this growth, called Logistic growth assumes that the growth rate G depends on the density of the population N as follows: G ( N ) = rN parenleftbigg K- N K parenrightbigg where r > 0 is a constant, called the intrinsic growth rate and K > 0 is a constant called the carrying capacity of the environment for the population. We will soon see that the largest population that would grow at all is N = K . We show a sketch of this function in Figure 6.1. But how would we arrive at such a sketch? We first answer the following questions: Find the population density that leads to the maximal growth rate. What is the maximal growth rate? For what population size is the growth rate zero? v.2005.1 - October 14, 2008 1 Math 102 Notes Chapter 6 Solution: G ( N ) = rN parenleftbigg K- N K parenrightbigg = rN- r K N 2 To find the maximal growth rate we differentiate G with respect to the variable N , remem- bering that K,r are here treated as constants. We get G prime ( N ) = r- 2 r K N. Setting G prime ( N ) = 0 and solving for N leads to r = 2 r K N so N = K 2 . By taking a second derivative we find that G primeprime ( N ) =- 2 r K which is negative for all population sizes. This tells us that the function G ( N ) is concave down, and that N = K/ 2 is a local maximum. Thus the density leading to largest growth rate is one half of the carrying capacity. The growth rate at this density is G ( K 2 ) = r parenleftbigg K 2 parenrightbigg parenleftBigg K- K 2 K parenrightBigg = r K 2 1 2 = rK 4 . To find the population size at which the growth rate is zero, we set G = 0 and solve for N : G ( N ) = rN parenleftbigg K- N K parenrightbigg = 0 . The two solutions are N = 0 (which is not very interesting, since when there is no population there is no growth) and N = K ....
View Full Document
This note was uploaded on 05/19/2011 for the course MATH 102 Math 102 taught by Professor Allard during the Winter '09 term at The University of British Columbia.
- Winter '09