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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

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                                              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 = 92/x 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
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
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 doesn’t 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
   Here’s 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 = 120/X 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
If L is fixed to be 8 + 2π, we have
                                       16 + 4π − 2w − πw
                                  h=                     ,
                                               4
so
                 w                         π                        1 π
              A=   (16 + 4π − 2w − πw) + w2 = (π + 4)w −              +      w2 .
                 4                         8                        2   8
                       π
So A = (π + 4)w − 1 +     , which is zero when
                       4
                                          π+4
                                     w=        = 4 ft.
                                          1+ π
                                             2

The dimensions are 4ft by 2ft.
Here’s another way to do it: We differentiate the L equation implicitly, keeping in mind it is
equal to a fixed constant:

                         dh      π             dh    1 + π/2    π+2
                   0=2      + 1+         =⇒       =−         =−     .
                         dw      2             dw       2        4
Now 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

    dA
So     = 0 when w = 2h, and this is true regardless of what L is fixed to be.
    dw
In our case, the relation w = 2h allows to solve L = 8 + 2π for h:

                          8 + 2π = 2h + (1 + π/2) (2h) = (4 + π)h

So h = 2 and w = 4.




                                           4

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Lesson 22: Optimization Problems (worksheet solutions)

  • 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 = 92/x 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 doesn’t 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 Here’s 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 = 120/X 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= , 4 so w π 1 π A= (16 + 4π − 2w − πw) + w2 = (π + 4)w − + w2 . 4 8 2 8 π So A = (π + 4)w − 1 + , which is zero when 4 π+4 w= = 4 ft. 1+ π 2 The dimensions are 4ft by 2ft. Here’s another way to do it: We differentiate the L equation implicitly, keeping in mind it is equal to a fixed constant: dh π dh 1 + π/2 π+2 0=2 + 1+ =⇒ =− =− . dw 2 dw 2 4 Now 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 dA So = 0 when w = 2h, and this is true regardless of what L is fixed to be. dw In our case, the relation w = 2h allows to solve L = 8 + 2π for h: 8 + 2π = 2h + (1 + π/2) (2h) = (4 + π)h So h = 2 and w = 4. 4