Name                                       MATH 115 MIDTERM III   Please work out each of the given problems.  Credit will be based on the steps that you show towards the final answer.  Show your work.   PROBLEM 1 Without the use of your graphing calculator, A)   Determine the relative extrema if any. B)    Determine where the function is increasing and decreasing. C)   Determine the inflection points if any. D)   Determine where the function is concave up and concave down.   E)  Use parts A through D to sketch the graph of the function   I.               y  =  x3 - 12x Solution We take the first and second derivatives for all four parts:         y'  =  3x2 - 12        y''  =  6x A)  Set the first derivative equal to zero.         3x2 - 12  =  0        3x2  =  12        x2  =  4         x  =  2        or        x  =  -2 Notice that the function is differentiable everywhere, hence there are no critical points other than 2 and -2. Now we use the second derivative test for x  =  -2:         y''(-2)  =  6(-2)  =  -12  <  0 Hence the graph has a relative maximum at x  =  -2.  Plugging into the original equation yields         (-2)3 - 12(-2)  =  16 Now we use the second derivative test for x  =  2:         y''(2)  =  6(2)  =  12  >  0 Hence the graph has a relative minimum at x  =  2.  Plugging into the original equation yields         (2)3 - 12(2)  =  -16 We conclude that (-2,16) is a relative maximum and (2,-16) is a relative minimum. B)  We want to find out where the first derivative is positive and where it is negative.  We draw a number line with the critical points -2 and 2 labeled.  We use test points to determine the sign of the first derivative.         y'(-3)  =  3(-3)2 - 12  =  15  > 0         y'(0)  =  3(0)2 - 12  =  -12  < 0         y'(3)  =  3(3)2 - 12  =  15  > 0         We can conclude that the graph is increasing at (-, -2) U (2, ) The graph is decreasing at (-2,2). C)  We set the second derivative equal to zero         6x  =  0        x  =  0 Plugging into the original equation yields         y  = 03 - 12(0)  =  0 Hence (0,0) is an inflection point. D)  We determine where the second derivative is positive and where it is negative. Since the inflection point occurs at x  =  0, we use test points on each side of 0.         y''(-1)  =  6(-1)  =  -6  < 0         y''(1)  =  6(1)  =  6  >  0         Hence the graph is concave down on (-, 0) and concave up on (0, ). E)  We use the information to sketch the graph.                                           II.            y  =  3x4 - 4x3 Solution We take the first and second derivatives for all four parts:         y'  =  12x3 - 12x2        y''  =  36x2 - 24x A)  Set the first derivative equal to zero.         12x3 - 12x2   =  0        12x2(x - 1)  =  0         x  =  0        or        x  =  1 Notice that the function is differentiable everywhere, hence there are no critical points other than 0 and 1. Now we use the second derivative test for x  =  0:         y''(0)  =  36(0)2 - 24(0)  =  0   The second derivative test fails, so we need to resort to the first derivative test.  We have          y'(-1)  =  12(-1)3 - 12(-1)2  =  -24  <  0         y'(.5)  =  12(.5)3 - 12(.5)2  =  -1.5  <  0 The first derivative test tells us that y(0) is neither a relative maximum nor a relative minimum. Plugging into the original equation yields         3(0)4 - 4(0)3  =  0 The point is (0,0) Now we use the second derivative test for x  =  1:         y''(1)  =  36(1)2 - 24(1)  =  12  >  0 Hence the graph has a relative minimum at x  =  1.  Plugging into the original equation yields         3(1)4 - 4(1)3  =  -1 We conclude that (0,0) is a horizontal point and (1,-1) is a relative minimum. B)  We want to find out where the first derivative is positive and where it is negative.  We draw a number line with the critical points 0 and 1 labeled.  We have already found out that y'(-1)  <  0 and y'(.5)  >  0.  We now test          y'(2)  =  12(2)3 - 12(2)2  =  24  > 0             We can conclude that the graph is decreasing on (-, 0) U (0,1) and is increasing on (1, ) C)  We set the second derivative equal to zero         36x2 - 24x  =  0        12x(3x - 2)  =  0         x  =  0        or         3x - 2  =  0         x  =  0        or        x  =  2/3 Plugging into the original equation yields         y  = 3(0)4 - 4(0)3  =  0 and         y  = 3(2/3)4 - 4(2/3)3  =  -16/27  =  -0.59 Hence (0,0) and (2/3,-0.59) are the inflection points. D)  We determine where the second derivative is positive and where it is negative.  We test         y''(-1)  =  36(-1)2 - 24(-1)  =  60  >  0         y''(.5)  =  36(.5)2 - 24(.5)  =  -3  <  0         y''(-1)  =  36(2)2 - 24(2)  =  96  >  0             We can conclude that the graph is concave up on (-, 0) U (2/3, ) and concave down on (0,2/3). E)  The graph is shown below                         III.                      3x - 4        y  =                                                     4x - 3                                   We take the first and second derivatives for all four parts.  We use the quotient rule                            (4x - 3)(3) - (3x - 4)(4)        y'  =                                                                                             (4x - 3)2                             12x - 9 - 12x + 16         =                                                                                     ( 4x - 3)2                               7         =                              =  7(4x - 3)-2                         ( 4x - 3)2  and to find the second derivative, we use the chain rule         u  =  4x - 3        u'  =  4         f(u)  =  7u -2        f '(u)  =  -14u -3 so         y''  =  (4)(-14u -3)  =  -56(4x - 3)-3                          -56  y''  =                                           (4x - 3)3    A)  Since the numerator of the first derivative is a constant, we see that it is never equal to zero.  There is a critical point at x  =  3/4 which is an asymptote.  We can conclude that there are no relative extrema. B)  We want to find out where the first derivative is positive and where it is negative.  We draw a number line with the critical point 3/4 labeled.  We use test points to determine the sign of the first derivative.         y'(0)  =   7(4(0) - 3)-2  =  7/9  >  0         y'(1)  =   7(4(1) - 3)-2  =  7  >  0             We can conclude that the graph is increasing for all x except at x = -3/4 where there is an asymptote. C)  Since the numerator of the second derivative is a constant, we see that it is never equal to zero.  We can conclude that there are no inflection points. D)  We want to find out where the second derivative is positive and where it is negative.  Although there are no inflection points, concavity can change at an asymptote.  Again, we draw a number line with the critical point 3/4 labeled.  We use test points to determine the sign of the second derivative.         y'(0)  =   -56(4(0) - 3)-3  =  -56/27  >  0         y'(1)  =   -56(4(1) - 3)-3  =  56  <  0 We conclude that the graph is concave up on (-, 3/4) and concave down on (3/4, ). E)  We use the above information to graph the function.                                               IV.                         x        y  =                                                   x2 + 1   We take the first and second derivatives for all four parts.  We use the quotient rule                            (x2 + 1)(1) - (x)(2x)        y'  =                                                                                             (x2 +1)2                             x2 + 1 - 2x2         =                                                                     (x2 + 1)2                               1 - x2          =                              =  (1 - x2)(x2 + 1)-2                         (x2 + 1)2  and to find the second derivative, we use the product and chain rule         y''  =  (1 - x2)[(x2 + 1)-2]' + (-2x)(x2 + 1)-2           u  =  x2 + 1        u'  =  2x         f(u)  =  u-2        f '(u)  =  -2u-3 so         y''  =  (1 - x2)[-4x(x2 + 1)-3] + (-2x)(x2 + 1)-2                       (1 - x2)(-4x)                  -2x         =                              +                                       (x2 + 1)3                 (x2 + 1)2                            4x3 - 4x                  -2x (x2 + 1)        =                              +                                             (x2 + 1)3                    (x2 + 1)3                            2x3 - 6x                 =                                             (x2 + 1)3        A)  We set the first derivative equal to zero.  This is the same as setting the numerator equal to zero.  We have         1 - x2  =  0        (1 + x)(1 - x)  =  0         x  =  -1    or        x  =  1   Use the second derivative test to find                                2(-1)3 - 6(-1)        y''(-1)  =                              =  1/2  >  0                                ((-1)2 + 1)3                                 2(1)3 - 6(1)        y''(1)  =                              =  -1/2  <  0                                ((1)2 + 1)3  Now plug -1 and 1 back into the original equation to get                                   1         y(1)  =                     =  1/2                              12 + 1                                    -1         y(-1)  =                       =  -1/2                              (-1)2 + 1 Notice that the denominator of the first derivative is never zero, hence there are no asymptotes.  We can conclude that there is a relative maximum at (-1,0.5) and a relative minimum at (1,0.5). B)  We want to find out where the first derivative is positive and where it is negative.  We draw a number line with the critical points -1 and 1 labeled.  We use test points to determine the sign of the first derivative.         y'(-2)  =   (1 - (-2)2)((-2)2 + 1)-2  =   -3/25  <  0           y'(0)  =   (1 - (0)2)((0)2 + 1)-2  =   1  >  0           y'(2)  =   (1 - (2)2)((2)2 + 1)-2  =   -3/25  <  0               Hence the graph is decreasing on (-, -1) U (1, ) and it is increasing on (-1,1). C)  We set the numerator of the second derivative equal to zero.  We have          2x3 - 6x  =  0        2x(x2 - 3)  =  0         x  =  0        x  =  -        x  =  We use a calculator to find the y coordinates to get inflection points at         (-1.7,-0.43), (0,0), (1.7,0.43) D)  We have already found the sign of the second derivative at 1 and -1.  We test two more points                                2(-2)3 - 6(-2)        y''(-2)  =                              =  -4/125  <  0                                ((-2)2 + 1)3                                 2(2)3 - 6(2)        y''(2)  =                              =  4/125  >  0                                ((2)2 + 1)3              E)  We use the information above to sketch the graph.                                                 PROBLEM 2  You manage the new Spockwood ski resort.  Your research shows that if you charge \$40 per lift ticket, you can expect to sell 3000 tickets.  For every \$5 increase in price you can expect to lose 250 customers.  Assume the demand equation is linear.   A)    Find the demand equation.   Solution Since the demand equation is linear, we need the equation of the line.  We find two points:         (3000,40)        and        (2750,45) The slope is                         45 - 40                     5                 1   m  =                              =                 =  -                                             2750 - 3000             -250              25 We have          p - 40  =  -1/25 (x - 3000)  =  -1/25 x  +  120 so that         p  =  -1/25 x  + 160                                   B)     What price will yield the maximum revenue?   Solution    Recall that revenue is equal to the number of units sold times the price.          R  =  xp  =  x (-1/25 x + 80)  =  -1/25 x2 + 160x To find the maximum revenue, we take the derivative and set it equal to zero         R'  =  -2/25 x  +  160   =   0         -2x  + 4000  =  0         x  =  2000 To be certain that this will give a maximum, we take the second derivative         R''  =  -2/25  This is always negative, hence it is negative at x  =  2000.  The second derivative test assures us that this point is a relative maximum. We plug this into the demand equation to find the price         p  =  -1/25 (2000) + 160  =  80 The price of \$80 will yield the maximum revenue.                                 C)     Is the demand elastic or inelastic when the price is \$45?     We compute                         p/x          h =                                        dp/dx First find x when p  =  45.  We have         45  =  -1/25 x +120        1/25 x  =  75         x  =  1875                           45 / 1875       h(45)  =                             =  - .6                                -1/25 Since the absolute value of this quantity is less than 1, we can conclude that the demand is in elastic.                               PROBLEM 3   At the blood bank, a cylindrical bottle without top is to contain 100 cc of blood.  What should the dimensions of the bottle be in order to minimize material costs?  Assume that the glass has the same thickness throughout.   Solution We want to minimize the surface area.  The surface area of the bottom (a circle) is       SAb  =  pr2  and the surface area of the rounded side is         SAs  =  2prh So that the total surface area is         SA  =  pr2 + 2prh The volume is given by         V  =  100  =  pr2 h so that                     100         h  =                                        pr2  Substituting into the surface area equation gives                                         100         SA  =  pr2 + 2pr                                                       pr2                                    200         SA  =  pr2 +                                                   r  We want to minimize the surface area, hence we take a derivative and set it equal to zero                                   200         SA'  =  2pr -                =  0                                     r2           2pr3 - 200  =  0            2pr3  =  200         r3  =  100/p  taking a cube root gives that r is approximately 3.17.  Notice that if we take a second derivative we get                                   400         SA''  =  2p +                                                     r3   Which is positive for r  =  3.17.  Hence this critical point yields a minimum as required.  We now find the value of h.  We have                         100         h  =                          =  3.17                      p(3.17)2  We construct the flask with a radius of 3.17 cm and a height of 3.17 cm.                                   PROBLEM 4  The department of fish and game have introduced a species of trout that was once native to the region.  The number (in hundreds) of this species t years after their introduction can be modeled by the equation                         90(2 + 5t)          P(t)  =                                                             1 + 10t Use differentials to approximate the change in the trout population from three to 3.5 years after their introduction. We have         DP  @  dp/dt Dt   Use the quotient rule to get the derivative                        (1 + 10t)(450)  -  90(2 + 5t)(10)          P'(t)  =                                                                                                               (1 + 10t)2  Now plug in in t  =  3.5 to get                             (1 + 10(3))(450)  -  90(2 + 5(3))(10)          P'(3.5)  =                                                                                  =  -1.4                                                               ( 1 + 10(3))2  We also have Dt  =  0.5  so that         DP  @  dp/dt Dt  =  -1.4(0.5)  =  -0.7 During this half year, the population decreases by approximately 70 fish.