# England_MAT_265_ONLINE_A_Spring_2020.Section_4.5.pdf -...

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This preview shows page 1 out of 4 pages. Unformatted text preview: England MAT 265 ONLINE A Spring 2020 Assignment Section 4.5 1. (1 point) Find two numbers A and B (with A ≤ B) whose difference is 38 and whose product is minimized. 3. (1 point) Find the length L and width W (with W ≤ L) of the rectangle with perimeter 100 that has maximum area, and then find the maximum area. L= A= B= Solution: We wish to maximize the product P = A · B. We want to express P as a function of just one variable, so we eliminate B by expressing it in terms of A. To do this we use the given information that the difference is 38. Thus W= Maximum area = Solution: We wish to maximize the area A = L ·W . We want to express A as a function of just one variable, so we eliminate W by expressing it in terms of L. To do this we use the given information that the perimeter is 100. Thus B − A = 38 From this equation we have B = A + 38, which gives P(A) = A(A + 38) = A2 + 38A The derivative is P0 (A) = 2A+38, so to find the critical numbers we solve the equation 2L + 2W = 100 2A + 38 = 0 P00 (A) = 2 > 0, which gives A = −19. Since minimum at A = −19 The corresponding value of B is From this equation we have W = 50 − L, which gives there is an absolute P(L) = L(50 − L) = 50L − L2 , 0 < L < 50 The derivative is P0 (L) = 50 − 2L, so to find the critical numbers we solve the equation B = −19 + 38 = 19 Correct Answers: • -19 • 19 50 − 2L = 0 2. (1 point) Find two positive numbers whose product is 81 and whose sum is a minimum. which gives L = 25. Since P00 (L) = −2 < 0, there is an absolute maximum at L = 25 Answer: , Solution: SOLUTION Let x and y be the two numbers. Since xy = 81, we have y = 81 x and we want to minimize The value of W is then W = 50 − 25 = 25 81 where x > 0 x 81 x2 − 81 f 0 (x) = 1 − 2 = x √ x2 The positive critical number is x = 81 = 9. Since f 0 (x) < 0 for 0 < x < 9 and f 0 (x) > 0 for x > 9, there is an absolute minimum at x = 9. Thus the numbers are 9 and 9. f (x) = x + and the corresponding maximum area is A = 25 · 25 = 625 Correct Answers: • 25 • 25 • 625 Correct Answers: • 9 • 9 1 4. (1 point) A fence is to be built to enclose a rectangular area of 290 square feet. The fence along three sides is to be made of material that costs 3 dollars per foot, and the material for the fourth side costs 13 dollars per foot. Find the length L and width W (with W ≤ L) of the enclosure that is most economical to construct. (c) Find the length L, width W , and height H of the resulting box that maximizes the volume. (Assume that W ≤ L). L= W= cm cm L= H= cm (d) The maximum volume of the box is cm3 . W= Solution: Let x and y denote the length and width of the rectangle and let y be the length of the side that costs 13 dollars per foot. The cost of the fence is then C = 3(2x) + 3y + 13y Solution: We want to express the cost as a function of just one variable, so we eliminate y by expressing it in terms of x. To do this we use the fact that the area enclosed is 290 square feet. Thus (a) The lenght of the box is L = 16 − 2x, the width is W = 6 − 2x and the height is H = x. Thus the volume is xy = 290 290 , which gives From this equation we have y = x 290 290 4640 C(x) = 6x + 3 + 13 = 6x + , x>0 x x x , so to find the critical numbers The derivative is C0 (x) = 6 − 4640 x2 we solve 4640 6− 2 = 0 x q V (x) = L ·W · H = (16 − 2x)(6 − 2x)x = 4x3 − 44x2 + 96x cm3 (b) Since the length, width and the volume must be positive, the domain is (0, 3). (c) The derivative is V 0 (x) = 12x2 − 88x + 96, so to find the critical numbers we need to solve the equation 12x2 − 88x + 96 = 0 √ 11 ± 49 . The roots are x = 3 √ 11 + 49 > 3, the only critical number in the domain is Since √3 11 − 49 x= . 3 √ 11 − 49 0 and V 0 (x) < 0 for Since V (x) > 0 for 0 < x < √ √ 3 11 − 49 11 − 49 <x<3, x= must give rise to an abso3 3 lute maximum in the interval (0, 6). Thus, the length, width and height that maximize the volume of the box are √ √ 11 − 49 26 + 2 11 L = 16 − 2 = ≈ 13.3333 3 3 √ √ 11 − 49 −4 + 2 11 W = 6−2 = ≈ 3.3333 3 3 √ 11 − 49 H= ≈ 1.3333 3 The solutions are x = ± 2320 . However, since x > 0, the only q 3 critical number is x = 2320 3 . Since C00 (x) = 13920 > 0 for x > 0, the critical number must give x3 rise to an absolute minimum in the domain (0, ∞). . The corresponding value of y is y = q290 2320 3 q Since y = q290 ≤ 2320 3 = x we have 2320 3 L= q 2320 3 290 and W = q 2320 3 Correct Answers: • 27.8088714861523 • 10.4283268073071 5. (1 point) A box is to be made out of a 6 cm by 16 cm piece of cardboard. Squares of side length x cm will be cut out of each corner, and then the ends and sides will be folded up to form a box with an open top. (a) Express the volume V of the box as a function of x. (d) The maximum volume is V= cm3 (b) Give the domain of V in interval notation. (Use the fact that length, width and volume must be positive.) V = L ·W · H ≈ 59.2593 Correct Answers: 2 • • • • • • But if (x, y) lies on the parabola, then x = y2 , so the expression for d becomes q d = y4 + (y − 3)2 (6 - 2 * x)(16 - 2 *x)x (0,3) 13.3333333333333 3.33333333333333 1.33333333333333 59.2592592592593 Instead of minimizing d, we minimize its square: d 2 = f (y) = y4 + (y − 3)2 6. (1 point) A rectangular storage container with an open top is to have a volume of 10 cubic meters. The length of its base is twice the width. Material for the base costs 14 dollars per square meter. Material for the sides costs 8 dollars per square meter. Find the cost of materials for the cheapest such container. Differentiating we obtain f 0 (y) = 4y3 + 2(y − 3) By inspection or graphically, we can see that the only real root is y = 1. Because of the geometric nature of the problem there must be an absolute minimum occurring when y = 1 since there is a closest point but not a farthest point. The minimum distance is p √ d = f (1) = 5 (Round to the nearest penny and Total cost = include monetary units. For example, if your answer is 1.095, enter \$1.10 including the dollar sign and second decimal place.) Solution: SOLUTION Let w and h denote the width of the base and the height of the container, respectively. Since the length of the base is twice the width, the volume of the container is Correct Answers: • 2.23606797749979 8. (1 point) A rectangle is inscribed with its base on the xaxis and its upper corners on the parabola y = 12 − x2 . What are the dimensions of such a rectangle with the greatest possible area? Width = Height = Solution: V = (2w)(w)h = 2w2 h = 10 so 10 2w2 The sides have total area 2(2w)h + 2wh = 6wh, while the base has area w(2w) = 2w2 , so the cost is h= 8 · (6wh) + 14 · (2w2 ) Substuting the expression for h gives 10 240 C(w) = 8 · 6w · 2 + 14 · (2w2 ) = + 28w2 2w w C0 (w) = − w= w>0 240 −240 + 56w3 + 56w = w2 w2  30 1/3 7 is the critical number. There is an absolute mini1/3 mum at this critical number since C0 (w) < 0 for 0 < w < 30 7 1/3 and C0 (w) > 0 for w > 30 . 7 The total minimum cost is    1/3 C 30 ≈ \$221.63 7 Correct Answers: • \\$221.63 (Click on graph to enlarge) The parabola is sketched in red in the figure and the inscribed rectangle is in blue. Let (x, y) be the vertex that lies in the first quadrant. Then the rectangle has sides of lengths 2x and y, so its area is 7. (1 point) Find the minimum distance from the parabola x − y2 = 0 to the point (0, 3). Minimum distance = Solution: SOLUTION The distance between the point (0, 3) and the point (x, y) is q d = x2 + (y − 3)2 A = 2xy To eliminate y we use the fact that (x, y) lies on the parabola y = 12 − x2 . Thus A(x) = 2x(12 − x2 ) = 24x − 2x3 3 • • • • • • But if (x, y) lies on the parabola, then x = y2 , so the expression for d becomes q d = y4 + (y − 3)2 (6 - 2 * x)(16 - 2 *x)x (0,3) 13.3333333333333 3.33333333333333 1.33333333333333 59.2592592592593 Instead of minimizing d, we minimize its square: d 2 = f (y) = y4 + (y − 3)2 6. (1 point) A rectangular storage container with an open top is to have a volume of 10 cubic meters. The length of its base is twice the width. Material for the base costs 14 dollars per square meter. Material for the sides costs 8 dollars per square meter. Find the cost of materials for the cheapest such container. Differentiating we obtain f 0 (y) = 4y3 + 2(y − 3) By inspection or graphically, we can see that the only real root is y = 1. Because of the geometric nature of the problem there must be an absolute minimum occurring when y = 1 since there is a closest point but not a farthest point. The minimum distance is p √ d = f (1) = 5 (Round to the nearest penny and Total cost = include monetary units. For example, if your answer is 1.095, enter \$1.10 including the dollar sign and second decimal place.) Solution: SOLUTION Let w and h denote the width of the base and the height of the container, respectively. Since the length of the base is twice the width, the volume of the container is Correct Answers: • 2.23606797749979 8. (1 point) A rectangle is inscribed with its base on the xaxis and its upper corners on the parabola y = 12 − x2 . What are the dimensions of such a rectangle with the greatest possible area? Width = Height = Solution: V = (2w)(w)h = 2w2 h = 10 so 10 2w2 The sides have total area 2(2w)h + 2wh = 6wh, while the base has area w(2w) = 2w2 , so the cost is h= 8 · (6wh) + 14 · (2w2 ) Substuting the expression for h gives 10 240 C(w) = 8 · 6w · 2 + 14 · (2w2 ) = + 28w2 2w w C0 (w) = − w= w>0 240 −240 + 56w3 + 56w = w2 w2  30 1/3 7 is the critical number. There is an absolute mini1/3 mum at this critical number since C0 (w) < 0 for 0 < w < 30 7 1/3 and C0 (w) > 0 for w > 30 . 7 The total minimum cost is    1/3 C 30 ≈ \$221.63 7 Correct Answers: • \\$221.63 (Click on graph to enlarge) The parabola is sketched in red in the figure and the inscribed rectangle is in blue. Let (x, y) be the vertex that lies in the first quadrant. Then the rectangle has sides of lengths 2x and y, so its area is 7. (1 point) Find the minimum distance from the parabola x − y2 = 0 to the point (0, 3). Minimum distance = Solution: SOLUTION The distance between the point (0, 3) and the point (x, y) is q d = x2 + (y − 3)2 A = 2xy To eliminate y we use the fact that (x, y) lies on the parabola y = 12 − x2 . Thus A(x) = 2x(12 − x2 ) = 24x − 2x3 3 ...
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