Arectangle has its base on the x-axis and its upper two vertices on the parabola y = 3

Find the smallest perimeter possible for a rectangle whose area is 25 in.2

Show that, among all rectangles with a given perimeter, the square has the largest area.

Arectangle has its base on the x-axis and its upper two vertices on the parabola y = 3 x2, as shown in Figure 5.56. What is the largest area the rectangle can have? 3 _ x2 _2 _1 2 1 4 5 x y Figure 5.56 The graph of y = 3 x2 together with the inscribed rectangle in Problem 3.

A rectangular study area is to be enclosed by a fence and divided into two equal parts, with the fence running along the division parallel to one of the sides. If the total area is 384 ft2, find the dimensions of the study area that will minimize the total length of the fence. How much fencing will be required?

A rectangular field is bounded on one side by a river and on the other three sides by a fence. Find the dimensions of the field that will maximize the enclosed area if the fence has a total length of 320 ft.

Find the largest possible area of a right triangle whose hypotenuse is 4 cm long.

Suppose that a and b are the side lengths in a right triangle whose hypotenuse is 5 cm long. What is the largest perimeter possible?

Suppose that a and b are the side lengths in a right triangle whose hypotenuse is 10 cm long. Show that the area of the triangle is largest when a = b.

A rectangle has its base on the x-axis, its lower left corner at (0, 0), and its upper right corner on the curve y = 1/x. What is the smallest perimeter the rectangle can have? 1

A rectangle has its base on the x-axis and its upper left and right corners on the curve y = _ 4 x2, as shown in Figure 5.57. The left and the right corners are equidistant from the vertical axis. What is the largest area the rectangle can have? _3 _2 _1 _0.5 2 0.5 1 2.5 3 x y 3 4 _ x2 Figure 5.57 The graph of y = (4 x2)1/2 together with the inscribed rectangle in Problem 10. 1

Denote by (x, y) a point on the straight line y = 43x. (See Figure 5.58.) x y _3 _2 _1 1 3 _6 _4 _2 4 2 6 8 10 12 14 f (x) (x, y) 4 _ 3x Figure 5.58 The graph of y = 4 3x in Problem 11. (a) Show that the distance from (x, y) to the origin is given by f (x) = _ x2 + (4 3x)2 (b) Give the coordinates of the point on the line y = 4 3x that is closest to the origin. (: Find x so that the distance you computed in (a) is minimized.) (c) Show that the of the distance between the point (x, y) on the line and the origin is given by g(x) = [ f (x)]2 = x2 + (4 3x)2 and find the minimum of g(x). Show that this minimum agrees with your answer in (b).

How close does the line y = 1 + 2x come to the origin?

How close does the curve y = 1/x come to the origin? (: Find the point on the curve that minimizes the of the distance between the origin and the point on the curve. If you use the square of the distance instead of the distance, you avoid dealing with square roots.)

How close does the circle with radius _ 2 and center (2, 2) come to the origin.

Show that if f (x) is a positive twice-differentiable function that has a local minimum at x = c, then g(x) = [f (x)]2 has a local minimum at x = c as well.

Show that if f (x) is a differentiable function with f (x) < 0 for all x R and with a local maximum at x = c, then g(x) = [ f (x)]2 has a local minimum at x = c.

Find the dimensions of a right circular cylindrical can (with bottom and top closed) that has a volume of 1 liter and that minimizes the amount of material used. (: One liter corresponds to 1000 cm3.)

Find the dimensions of a right circular cylinder that is open on the top, is closed on the bottom, holds 1 liter, and uses the least amount of material.

A circular sector with radius r and angle has area A. Find r and so that the perimeter is smallest when (a) A = 2 and (b) A = 10. (: A = 1 2 r 2, and the length of the arc s = r, when is measured in radians; see Figure 5.59.) u s r Figure 5.59 The circular sector in Problems 19 and 20.

A circular sector with radius r and angle has area A. Find r and so that the perimeter is smallest for a given area A. (: A = 1 2 r 2, and the length of the arc s = r, when is measured in radians; see Figure 5.59.)

Repeat Example 4 under the assumption that the top of the can is made out of aluminum that is three times as thick as the aluminum used for the wall and the bottom.

Find two positive numbers a and b such that a + b = 20 and ab is a maximum.

Find two numbers a and b such that a b = 4 and ab is a minimum.

Classical Model of Viability Selection Consider a population of diploid organisms (i.e., each individual carries two copies of each chromosome). Genes reside on chromosomes, and we call the location of a gene on a chromosome a locus. Different versions of the same gene are called alleles. Let us examine the case of one locus with two possible alleles, A1 and A2. Since the individuals are diploid, the following types, called genotypes, may occur: A1A1, A1A2, and A2A2 (where A1A2 and A2A1 are considered to be equivalent). If two parents mate and produce an offspring, the offspring receives one gene from each parent. If mating is random, then we can imagine all genes being put into one big gene pool from which we choose two genes at random. If we assume that the frequency of A1 in the population is p and the frequency of A2 is q = 1 p, then the combination A1A1 is picked with probability p2, the combination A1A2 with probability 2pq (the factor 2 appears because A1 can come from either the father or the mother), and the combination A2A2 with probability q2. We assume that the survival chances of offspring depend on their genotypes. We define the quantities w11, w12, and w22 to describe the differential survival chances of the types A1A1, A1A2, and A2A2, respectively. The ratio A1A1:A1A2:A2A2 among adults is given by p2w11:2pqw12 :q2w22 The average fitness of this population is defined as w = p2w11 + 2pqw12 + q2w22 We will investigate the preceding function. Since q = 1 p, w is a function of p only; specifically, w(p) = p2w11 + 2p(1 p)w12 + (1 p)2w22 for 0 p 1.We consider the following three cases: (i) Directional selection: w11 > w12 > w22 (ii) Overdominance: w12 > w11,w22 (iii) Underdominance: w12 < w11,w22 (a) Show that w(p) = p2(w11 2w12 + w22) + 2p(w12 w22) + w22 and graph w(p) for each of the three cases, where we choose the parameters as follows: (i) w11 = 1,w12 = 0.7,w22 = 0.3 (ii) w11 = 0.7,w12 = 1,w22 = 0.3 (iii) w11 = 1,w12 = 0.3,w22 = 0.7 (b) Show that dw dp = 2p(w11 2w12 + w22) + 2(w12 w22) (c) Find the global maximum of w(p) in each of the three cases considered in (a). (Note that the global maximum may occur at the boundary of the domain of w.) (d) We can show that under a certain mating scheme the gene frequencies change until w reaches its global maximum. Assume that this is the case, and state what the equilibrium frequency will be for each of the three cases considered in (a).

Continuation of Problem 94 from Section 4.3 We discussed the properties of hatching offspring per unit time, w(t), in the species . The function w(t) was given by w(t) = f (t) C + t where f (t) is the proportion of offspring that survive if t is the time spent brooding and where C is the cost associated with the time spent searching for other mates. We assume now that f (t), t 0, is twice differentiable and concave down with f (0) = 0 and 0 f 1. The optimal brooding time is defined as the time that maximizes w(t). (a) Show that the optimal brooding time can be obtained by finding the point on the curve f (t) where the line through (C, 0) is tangential to the curve f (t). (b) Use the procedure in (a) to find the optimal brooding time for f (t) = t 1+t and C = 2. Determine the equation of the line through (2, 0) that is tangential to the curve f (t) = t 1+t , and graph both f (t) and the tangent together.

Optimal Age of Reproduction Semelparous organisms breed only once during their lifetime. Examples of this type of reproduction can be found in Pacific salmon and bamboo. The per capita rate of increase, r , can be thought of as a measure of reproductive fitness. The greater the value of r , the more offspring an individual produces. The intrinsic rate of increase is typically a function of age x. Models for agestructured populations of semelparous organisms predict that the intrinsic rate of increase as a function of x is given by r (x) = ln [l(x)m(x)] x where l(x) is the probability of surviving to age x and m(x) is the number of female offspring at age x. The optimal age of reproduction is the age x that maximizes r (x). (a) Find the optimal age of reproduction for l(x) = eax and m(x) = bxc where a, b, and c are positive constants. (b) Use a graphing calculator to sketch the graph of r (x) when a = 0.1, b = 4, and c = 0.9.

Optimal Age at First Reproduction Iteroparous organisms breed more than once during their lifetime. Consider a model in which the intrinsic rate of increase, r , depends on the age of first reproduction, denoted by x, and satisfies the equation ex(r (x)+L)(1 ekx )3c 1 e(r (x)+L) = 1 (5.13) where k, L, and c are positive constants describing the life history of the organism. The optimal age of first reproduction is the age x for which r (x) is maximized. Since we cannot separate r (x) in the preceding equation, we must use implicit differentiation to find a candidate for the optimal age of reproduction. (a) Find an equation for dr dx . [: Take logarithms of both sides of (5.13) before differentiating with respect to x.] (b) Set dr dx = 0 and show that this gives r (x) = 3kekx 1 ekx L [To find the candidate for the optimal age x, you would need to substitute for r (x) in (5.13) and solve the equation numerically. Then you would still need to check that this solution actually gives you the absolute maximum. It can, in fact, be done.]