MTH 1125 (10 am class) Test #3 - Solutions Fall 2013

Transcription

1 MTH 1125 (10 am class) Test # - Solutions Fall 201 Pat Rossi Name Show CLEARLY how you arrive at your answers!!! 1. ( ) = on the interval [ ] ; Find the absolute maximum and Absolute minimum vales of ( ) Since ( ) is 1 continuous (it s a polynomial) on the 2 closed, finite interval [ ] we can use the Absolute Max/Min Value Test i. Compute 0 ( ) and find the critical numbers 0 ( ) = a. Type a ( 0 ( ) =0) Set 0 ( ) = = = =0 ( +2)( +1)=0 = 2; = 1 type a crit. numbers b. Type b ( 0 ( ) undefined) There are none. ii. Plug critical numbers and endpoints into the original function. ( ) = 2 ( ) +9( ) 2 +12( ) + = 6 Abs Min Value ( 2) = 2 ( 2) +9( 2) 2 +12( 2) + = 1 ( 1) = 2 ( 1) +9( 1) 2 +12( 1) + = 2 () = 2 () +9() () + = 174 Abs Max Value Abs Max Value = 174 (attained at =) Abs Min Value = 6 (attained at = )

2 = 2 2 Compute 0 i. Differentiate both sides w.r.t = {z} 2 {z} = 1 st 2 ii. Solve algebraically for 0 2 nd {z} {z} {z } 1 st prime 2 nd 2 nd prime 2 {z} 1 st a. Get 0 terms on left side, all other terms on right side =6 2 2 b. Factor out =6 2 2 c. Divide both sides by the cofactor of 0 0 = = = =

3 . ( ) = Determine the intervals on which ( ) is increasing/decreasing, and identify all relative maximums and minimums. i. Compute 0 ( ) and find critical numbers 0 ( ) = a. Type a ( 0 ( ) =0) Set 0 ( ) = = = =0 ( +2)( 1) = 0 = 2; =1critical numbers b. Type b ( 0 ( ) undefined) There are none. ii. Draw a sign graph of 0 ( ) using the critical numbers to partition the -axis iii. From each interval select a test point to plug into 0 ( )!2 1 fn(x) = 6x 2 + 6x! 12! test pt 0 test pt 2 test pt !!!!! fn(x) > 0 fn(x) < 0 fn(x) > 0 ( ) is increasing on the intervals ( 2) and (1 ) (Because 0 ( ) is positive on these intervals) ( ) is decreasing on the interval ( 2 1) (Because 0 ( ) is negative on this interval)

4 iv. To find the relative maxes and mins, sketch a rough graph of ( ) Rel Max (!2, f(!2) ) x =!2 x = 1 (1, f(1) ) Rel Min ( ) is increasing on the intervals ( 2) and (1 ) ( ) is decreasing on the interval ( 2 1) ( 2 ( 2)) = ( 2 2) Relative Max (1 (1)) = (1 4) Relative Min 4

5 4. ( ) = Determine the intervals on which ( ) is concave up/concave down, and identify all points of inflection. i. Compute 00 ( ) and find possible points of inflection 0 ( ) = ( ) = a. Type a ( 00 ( ) =0) Set 00 ( ) = = = =0 ( +)( 2) = 0 = ; and =2possible points of inflection b. Type b ( 00 ( ) undefined) There are none. ii. Draw a sign graph of 00 ( ) using the possible points of inflection to partition the -axis iii. From each interval select a test point to plug into 00 ( )! 2 fn(x) = 12x x! 72!4 test pt 0 test pt test pt !!!!! fo(x) > 0 fo(x) < 0 fo(x) > 0 ( ) is concave up on the intervals ( ) and (2 ) (Because 00 ( ) 0 on these intervals) ( ) is concave down on the interval ( 2) (Because 00 ( ) 0 on this interval) Since ( ) changes concavity at = and =2 the points: ( ( )) = ( 15) and (2 (2)) = (2 105) are points of inflection. 5

6 5. Given the following information, graph ( ) 0 ( ) 0 on ( ) 0 ( ) 0 on ( 0) 00 ( ) 0 on ( ) lim ( ) = 0 ( ) 0 on (0 ) 00 ( ) 0 on ( ) lim ( ) = 0 ( ) 0 on ( ) 00 ( ) 0 on ( ) To draw a graph of ( ) let s determine what this information tells us. 0 ( ) 0 on ( ) ( ) is increasing on ( ) 0 ( ) 0 on ( 0) ( ) is decreasing on ( 0) 0 ( ) 0 on (0 ) ( ) is increasing on (0 ) 0 ( ) 0 on ( ) ( ) is decreasing on ( ) 00 ( ) 0 on ( ) ( ) is concave down on the interval ( ) 00 ( ) 0 on ( ) ( ) is concave up on the interval ( ) 00 ( ) 0 on ( ) ( ) is concave down on the interval ( ) lim ( ) = as = ( ) goes to lim ( ) = as = ( ) goes to Sketch a graph of ( ) using this information. Increasing Concave Down Decreasing Concave Up Increasing Concave Up Decreasing Concave Down x =! x = x = 0 6

7 6. Farmer Joe has 1000 yards of fencing with which he will construct a rectangular pen. One side of the pen will border on a straight canal, and no fencing will be required on that side. In addition, Farmer Joe will use some of the fencing to partition the pen into three smaller pens of equal size and similar shape (See picture below). What should the overall dimensions of the pen be in order for the pen to contain the largest area possible? canal canal no fence canal y y y y x x x i. Determine the quantity to be maximized/minimized - give it a name, Maximize the overall area of the pen, = ii. Express as a function of one variable. (Refer to a restriction stated in the problem to do this) Restriction: Farmer Joe will use exactly 1000 yards of fencing. Since the fencing consists of three segments of length and four segments of length we have: +4 = 1000 yds 4 = 1000 yds = 250 yds 4 Substituting this into the equation = we have: = 250 yds 4 = 750 yds i.e., ( ) = 750 yds iii. Determine the restrictions on 0 yds 1000 yds iv. Maximize/minimize, using the techniques of calculus. Observe: ( ) is 1 continuous (it s a polynomial) on the 2 closed, finite interval yds yds Therefore, we can use the Absolute Max/Min Value Test 7

8 1. Compute 0 ( ) and find the critical numbers 0 ( ) = 750 yds 9 2 a. Type a ( 0 ( ) =0) 0 ( ) = 750 yds 9 2 =0 750 yds 9 2 =0 9 2 = 750 yds = 500 yds - critical number b. Type b ( 0 ( ) is undefined) There are none. 2. Plug the critical numbers and endpoints into the original function. (0 yds) = 750 (0 yds) 9 4 (0 yds)2 =0yds yds = yds yds = yds 9 4 yds 2 = yds 2 Abs Max Value 1000 yds 2 =0yds 2 5. Make sure that we ve solved the original question (problem). What should the overall dimensions... be We have the Abs Max Area when = 500 yds Length = x = 500 yds =500yds Width = = 250 yds 4 = 250 yds yds = 125 yds Length = 500 yds Width = 125 yds 8