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Need a tutor? Click this link and get your first session free! * AP® is a trademark registered and owned by the College Board, which was not involved in the production of, and does not endorse, this site. Mobile Notice You appear to be on a device with a narrow screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width. 1. Solutions to Worksheet for Section 4.5 Optimization Problems V63.0121, Calculus I Summer 2010 1. An advertisement consists of a rectangular printed region plus 1 in margins on the sides and 2 in margins on the top and bottom. If the area of the printed region is to be 92 in2 , ﬁnd the dimensions of the printed region and overall advertisement that minimize the total area. Solution. The objective is to minimize the total area while the printed area is ﬁxed. 2 cm Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis 1 cm 1 cm y nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu 2 cm x Let the dimensions of the printed region be x and y; then the printed area is P = xy ≡ 92 while the total area is A = (x + 2)(y + 4). From the P equation we can isolate y = 92x and substitute it into the A equation to get 92 FOIL 184 184 A(x) = (x + 2) +4 = 92 + 4x + + 8 = 100 + 4x + x x x 1 2. The domain of A is (0, ∞). We want to ﬁnd the absolute minimum value of A. Taking derivatives, dA 184 =4− 2 dx x dA √ The equation = 0 gives a single critical point when x = 46 (the other possible one at √ dx − 46 is outside the domain). Taking another derivative, d2 A 368 = 3 dx2 x √ which is positive along the entire domain of A. So the single local minimum at x =√ 46 is in fact the global minimum. Therefore, the dimensions of the printed region are x = 46 in by 92 √ √ √ y = √ = 2 46 in. The dimensions of the paper are 46 + 2 in × 2 46 + 4 in. 46 2. An advertisement consists of a rectangular printed region plus 1 in margins on the sides and 1.5 in margins on the top and bottom. If the total area of the advertisement is to be 120 in2 , what dimensions should the advertisement be to maximize the area of the printed region? Solution. Using the notation of the previous problem, the objective is to maximize P while keeping A ﬁxed. 2 cm Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor inci- didunt ut labore et dolore magna aliqua. 1 cm 1 cm y Ut enim ad minim ve- niam, quis nostrud ex- ercitation ullamco la- boris nisi ut aliquip ex ea commodo conse- quat. Duis aute irure 2 cm x 120 In this problem, P = xy, but A = (x + 2)(y + 3). Isolating y in A ≡ 120 gives y = −3 x+2 which yields 120 120x P =x −3 = − 3x x+2 x+2 2 3. The domain of P is again (0, ∞). We want to ﬁnd the absolute maximum value of P . Taking derivatives, dP (x + 2)(120) − (120x)(1) 240 − 3(x + 2)2 = 2 −3= dx (x + 2) (x + 2)2 There is a single critical point when √ (x + 2)2 = 80 =⇒ x = 4 5 − 2 (again, the negative critical point doesnt count). The second derivative is d2 P −480 2 = dx (x + 2)3 √ which is negative all along the domain of P . Hence the unique critical point x = 4 5 − 2 cm √ is the absolute maximum of P . This means the paper width is 4 5 cm, and the paper length 120 √ is √ = 6 5 cm. 4 5 Heres another way to do it: Let X and Y be the width and height of the advertisement. Then we need to maximize P = (X − 2)(Y − 3) subject to the constraint that A = XY ≡ 120. Substituting Y = 120X into P gives 120 240 P = (X − 2) −3 = 126 − − 3X X X and we need to ﬁnd the maximum value of P over (0, ∞). 3. A Norman window has the outline of a semicircle on top of a rectangle. Suppose there is 8 + 2π feet of wood trim available. Find the dimensions of the rectangle and semicircle that will maximize the area of the window. Solution. Let h and w be the height and width of the window. We have π L = 2h + w + w 2 π w 2 A = wh + 2 2 3 4. If L is ﬁxed to be 8 + 2π, we have 16 + 4π − 2w − πw h= , 4so w π 1 π A= (16 + 4π − 2w − πw) + w2 = (π + 4)w − + w2 . 4 8 2 8 πSo A = (π + 4)w − 1 + , which is zero when 4 π+4 w= = 4 ft. 1+ π 2 The dimensions are 4ft by 2ft.Heres another way to do it: We diﬀerentiate the L equation implicitly, keeping in mind it isequal to a ﬁxed constant: dh π dh 1 + π2 π+2 0=2 + 1+ =⇒ =− =− . dw 2 dw 2 4Now diﬀerentiate A treating h as a function of w: dA dh π π+2 π w =h+w + w =h−w + w =h− dw dw 4 4 4 2 dASo = 0 when w = 2h, and this is true regardless of what L is ﬁxed to be. dwIn our case, the relation w = 2h allows to solve L = 8 + 2π for h: 8 + 2π = 2h + (1 + π2) (2h) = (4 + π)h So h = 2 and w = 4. 4

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