Applied Maximum and Minimum Problems
16.31 Find the maximum area of any rectangle which may be inscribed in a circle of radius 1
I Let the center of the circle be the origin. We may assume that the sides of the rectangle are parallel to the coordinate axes. Let 2x be the length of the horizontal sides and 2y be the length of the vertical sides. Then x2 + v2 = l, so 2x + 2 y D v = 0, D v = -xly. The area A = (2x)(2v) = 4xv. So, DA =
Solving DfA=0, we find y — x, 2x2 = l, Since the second derivative is negative, the unique critical number vields an absolute maximum. The maximum area is
16.32 A factory producing a certain type of electronic component has fixed costs of $250 per day and variable costs of 90*, where x is the number of components produced per day. The demand function for these components is p(x) = 250 - x, and the feasible production levels satisfy 0 & x < 90. Find the level of production for maximum profit.
Download from Wow! eBook <www.wowebook.com>
APPLIED MAXIMUM AND MINIMUM PROBLEMS D 125
16.33
16.34
16.35
16.36
I The daily income is *(250-*), since 250-* is the price at which x units are sold. The profit G = *(250 - x) - (250 + 90*) = 250x - x2 - 250 - 90x = 160* - x2 - 250. Hence, DXG = 160 -2x, D2XG = -2. Solving DXG = 0, we find the critical number x = 80. Since the second derivative is negative, the unique critical number yields an absolute maximum. Notice that this maximum, taken over a continuous variable x, is assumed for the integral value x = 80. So it certainly has to remain the maximum when x is restricted to integral values (whole numbers of electronic components).
A gasoline station selling x gallons of fuel per month has fixed cost of $2500 and variable costs of 0.90*. The demand function is 1.50 — 0.00002* and the station's capacity allows no more than 20,000 gallons to be sold per month. Find the maximum profit.
I The price that x gallons can be sold at is the value of the demand function. Hence, the total income is *(1.50- 0.00002*), and the profit G = *(1.50- 0.00002*) - 2500- 0.90* = 0.60* - 0.00002*2 - 2500.
Hence, DXG = 0.60 - 0.00004*, and D 2G = -0.00004. Solving DXG = 0, we find 0.60 = 0.00004*, 60,000 = 4*, x = 15,000. Since the second derivative is negative, the unique critical number * = 15,000 yields the maximum profit $2000.
Maximize the volume of a box, open at the top, which has a square base and which is composed of 600 square inches of material.
I Let s be the side of the base and h be the height. Then V=s2h. We are told that 600 = s2 + 4/w.
Hence, h = (600-s2)/4s. So V=s2[(600-s2)/4s] = (s/4)(600- s2) = 150s - Js3. Then DsV=150-i*2, D2V=-|i. Solving DSV=0, we find 200 = s2, 10V2 = s. Since the second derivative is negative, this unique critical number yields an absolute maximum. When s = 10V2, h — 5V2.
A rectangular garden is to be completely fenced in, with one side of the garden adjoining a neighbor's yard. The neighbor has agreed to pay for half of the section of the fence that separates the plots. If the garden is to contain 432 ft2, find the dimensions that minimize the cost of the fence to the garden's owner.
f Let y be the length of the side adjoining the neighbor, and let * be the other dimension. Then 432 = *y, Q = xD,y + y, Dxy = -y/x. The cost C = 2x + y + \y = 2* + |y. Then, DfC = 2+\(Dfy) = 2+|(-y/*) and £>2C = -\(xDxy -y)/x2 = -\(-y - y)/x2 = 3y/*2. Setting DXC = 0, We obtain 2 = 3y/2*, 4x = 3y, y = f*, 432 = *(4*/3), 324 = *2, * = 18, y = 24. Since the second derivative is posi-tive, the unique critical number * = 18 yields the absolute minimum cost.
A rectangular box with open top is to be formed from a rectangular piece of cardboard which is 3 inches x 8 inches. What size square should be cut from each corner to form the box with maximum volume? (The cardboard is folded along the dotted lines to form the box.)
Fig. 16-11
I Let * be the side of the square that is cut out. The length will be 8-2*, the width 3-2*, and the height *. Hence, the volume V= *(3 - 2*)(8 - 2*); so DXV= (1)(3 - 2*)(8 - 2*) + *(-2)(8 - 2*)
*(3-2*)(-2) = 4(3*-2)(*-3), and D2V=24*-44. Setting DXV=0, we find * = | or x = 3.
Since the width 3 of the cardboard is greater than 2*, we must have *<§. Hence, the value * = 3 is impossible. Thus, we have a unique critical number * = I , and, for that value, the second derivative turns out to be negative. Hence, that critical number determines an absolute maximum for the volume.
16.37 Refer to Fig. 16-12. At 9 a.m., ship B was 65 miles due east of another ship, A. Ship B was then sailing due west at 10 miles per hour, and A was sailing due south at 15 miles per hour. If they continue their respective courses, when will they be nearest one another?
126 0 CHAPTER 16
I Let the time / be measured in hours from 9 a.m. Choose a coordinate system with B moving along the jt-axis and A moving along the _y-axis. Then the ^-coordinate of B is 65 - lOt, and the y-coordinate of B is -15f.
Let u be the distance between the ships. Then u2 = (\5t)2 + (65 - IQt)2. It suffices to minimize u2. D,(u2) = 2(150(15) + 2(65 - I0t)(-10) = 650f- 1300, and D,2(«2) = 650. Setting D,(w2) = 0, we obtain t = 2, Since the second derivative is positive, the unique critical number yields an absolute minimum. Hence, the ships will be closest at 11 a.m.
Fig. 16-12 Fig. 16-13
16.39 A wall 8 feet high is 3.375 feet from a house. Find the shortest ladder that will reach from the ground to the house when leaning over the wall.
I Let x be the distance from the foot of the ladder to the wall. Let y be the height above the ground of the point where the ladder touches the house. Let L be the length of the ladder. Then L2 = (x + 3.37S)2 + y2. It suffices to minimize L2 By similar triangles, y/8 = (x + 3.375)Ix. Then Dxy = -27/x2. Now, DX(L2) = 2(x + 3.375) + 2yDty = 2(.v + 3.375) + 2(8/x)(x + 3.375)(-27Ix2) = 2(x + 3.375)(1 - 216/*3). Solv-ing Dr(L2) = 0, we find the unique positive critical number x = 6. Calculation of £>2(L2) yields 2 + (2/*4)[(27)2 + yx] > 0. Hence, the unique positive critical number yields the minimum length. When x = 6, y = f , L =-^ = 15.625 ft.
Fig. 16-14
16.40 A company offers the following schedule of charges: $30 per thousand for orders of 50,000 or less, with the charge per thousand decreased by 37.5 cents for each thousand above 50,000. Find the order that will maximize the company's income.
I Let x be the number of orders in thousands. Then the price per thousand is 30 for x s 50 and 30-j|(jt-50) for *>50. Hence, for *<50, the income 7 = 30*, and, for jt>50, / = ;t[30-g(;t-50)]= ™x - Ix2. So, for x<50, the maximum income is 1500 thousand. For *>50, DJ = ir ~ !* and D2/ = — | . Solving DXI = 0, A: = 65. Since the second derivative is negative, x = 65 yields the maximum income for x > 50. That maximum is 3084.375 thousand. Hence, the maximum income is achieved when 65,000 orders are received.
25 + x , 3x2 = 25, x = 5V3/3 = 2.89. Since the second derivative is positive, the unique critical number yields the absolute minimum time.
we obtain
and the distance walked is Hence the total time
Let x be the distance between A and the landing point. Then the distance rowed is
16.38 A woman in a rowboat at P, 5 miles from the nearest point A on a straight shore, wishes to reach a point B, 6 miles from A along the shore (Fig. 16-13). If she wishes to reach B in the shortest time, where should she land if she can row 2 mi/h and walk 4 mi/h?
Then Setting
APPLIED MAXIMUM AND MINIMUM PROBLEMS 0 127 16.41 A rectangle is inscribed in the ellipse *2/400 + y2/225 = 1 with its sides parallel to the axes of the ellipse (Fig.
16-15). Find the dimensions of the rectangle of maximum perimeter which can be so inscribed.
I x/200 + (2y/225)Dsy = 0, Dxy = -(9x/16y). The perimeter P = 4x+4y, so DxP = 4 + 4Dxy = 4(l-9*/16.y) = 4(16y-9;t)/16.y and D\P= -\\y -x(-9x!16y)]/y2 = -?(16/ + 9*2)/16/<0. Solving DrP = 0, I6y = 9x and, then, substituting in the equation of the ellipse, we find x2 = 256, x = 16, y = 9.
Since the second derivative is negative, this unique critical number yields the maximum perimeter.
Fig. 16-15 Fig. 16-16
16.42
16.43
ber >b is y = 3b, and, by the first derivative test, this yields a relative minimum, which, by the uniqueness of the critical number, must be an absolute minimum.
Find the dimensions of the right circular cylinder of maximum volume that can be inscribed in a right circular cone of radius R and height H (Fig. 16-17).
I Let r and h be the radius and height of the cylinder. By similar triangles, r/(H-h) = R/H, r = (RIH)(H-h). The volume of the cylinder V= Trr2h = ir(R2/H2)(H- h)2h. Then Dl,V=(trR2/H2)(H -h)(H — 3h), so the only critical number for h < H is h = H/3. By the first-derivative test, this yields a relative maximum, which, by the uniqueness of the critical number, is an absolute maximum. The radius r =
16.44 A rectangular yard must be enclosed by a fence and then divided into two yards by a fence parallel to one of the