M119 Fall 06 Midterm solutions Prepared by Jeremy Boggess. Please email questions/errors to [email protected] Form A (Problem # 5 is the tobacco problem). 1) C

C (q ) = 5000 + 3q R(q ) = pq = 7 q Breakeven is when Revenue = Cost

or when Profit =0

" (q ) = R ! C = 7 q ! (5000 + 3q ) " (q ) = 4q ! 5000 4q = 5000 q = 1250 tickets

7 q = 5000 + 3q 4q = 5000 q = 1250 tickets

2) A A) This is LOOKS like the formula for a power function, but a power function must have an x in the base and constant in the exponent. The correct derivative would be: Example:

d x c = ln(c)c x dx

d x 5 = ln(5)5 x dx

B) This is the correct formula for an exponential function. Ex.

d (5 x ) e = e (5 x ) * (5) by the chain rule dx

C) This is the correct formula for a power function. Ex.

d 5 x = 5 x (5!1) dx

D) This is the correct formula for a power function. Ex.

d (5 ln( x) )= 5 * 1 = 5 dx x x

3) A For anything growing at a continuous rate, you should use Plug in to the answers to see which are correct: So

A)

P = P0 e rt or in this case P = P0 e.05t

P' = P0 e.05t (.05)

P(2) = P0 e.05( 2 )

P' (2) = P0 e.05( 2 ) (.05)

P(3) = P0 e.05(3)

P' (3) = P0 e.05(3) (.05)

P0 e.05(3) (.05) = .05( P0 e.05(3) ) (true)

P0 e.05(3) = k ( P0 e.05( 2 ) ) B)

(false, but they are relatively close) P0 e.05(3) = k = e.05 ! 1.051271 .05 ( 2 ) P0 e

C)

P' (3) = P0 e.05(3) (.05) (false)

D)

P0 e.05(3) (.05) ! 1.05( P0 e.05( 2 ) )(.05) (false, almost like B)

4) E

32 ! ! 64 96 2 = = Since the table is linear, then m = 96 ! ! 48 144 3 y = y 2 + m( x ! x 2 ) 2 2 2 y = !64 + ( x ! ! 48) = !64 + x + 32 = x ! 32 3 3 3 So the y-intercept is (0,-32) (not listed) 5) A

m=

"P 1787 ! 1517 270 million lbs = = = 135 million lbs per year "t 1996 ! 1994 2 years

6) B

P(t ) = 98000(.84) t P(t ) = P0 (1 + r ) t

so

1 + r = .84 r = .84 ! 1 = !0.16

which means it is decreasing (due to the negative slope) at an annual rate of 16%. 7) A 12 t

r $ ' A = P0 %1 + " & 12 #

12*5

' .0425 $ A = 10000%1 + " 12 # &

! $12363.01899

8) C

B = Pe B = e rt P

B ln( ) = rt p B ln( ) P =t r

rt

or

1 &B# ln$ ! r %P"

9) B Since it doesn’t ask for a continuous rate, you want the r in By either plugging in or using 1

y = Pa t = P(1 + r ) t and not y = Pe rt .

y2 = a x2 ! x1 , you find that 190205 = a10 so y1 174989

1 + r ! 1.00837279 ' 190205 $ 10 % " = a ! 1.00837279 . Substitution in 1 + r you get r ! .00837279 ! 0.837279% & 174989 #

10) B

C = k * O2 8 = k (10) 2 C = .08(20) 2 plugging in O = 20 we get: 8 C = .08(400) = $32 million = .08 = k 100 C = .08 * O 2 11) C The new graph is flipped vertically (multiply by a negative on the outside of the function) and is shifted left 2 (replace (x) with (x+2) as all x operations are the inverses of your intuition). The other choices: A) flips horizontally and then up 2 B) right 2 C) OK D) Flips vertically and then down 2 12) C

f ( x) = 5000 x100 is an even shape (like a parabola) with both tails going up. g ( x) = e.01x is exponential growth (which goes from very close to zero to VERY large, always larger than any power function). So as x # !" (negative numbers that are large enough in absolute value), then f is going to infinity and g is going to 0, hence f ( x) > g ( x) . Likewise if x " ! (positive numbers that are large enough in absolute value), then f is going to infinity but g is going to infinity faster), hence f ( x) < g ( x) . Therefore both (A) and (B) are correct ! (D) Both A and B are true. Many students noticed that if just A were true, then E would be true and likewise if only B were true, then E would also be true. In other words, E would also be an acceptable answer if A or B exclusively was an acceptable answer. 13) C Since f is concave up, we know that f’ is increasing, so you could go immediately to (C). You could also try to estimate the slopes visually by drawing in the tangent lines and using rise over run to estimate the derivatives. Rough estimates are:

f ' (1) " !4

f ' (2) " !2

f ' (3) ! 0

f ' (4) ! 2

14) C I recommend changing each statement to a statement about the graph, N

dN = N ' increasing ! N concave up dt d 2N = N ' ' negative ! N concave down B) dt 2 d 2N = N ' ' positive ! N concave up C) dt 2 dN = N ' increasing ! N concave up D) dt A)

Now checking the intervals, we find that the only match is (C) N concave up from 1950-1970.

15) A

f ( x ) " y 2 + m( x ! x 2 )

f ( x + #x) " f ( x) + f ' ( x)#x f (20 + ! 2) " f (20) + f ' (20)( ! 2) = 68 + (!.025)(!2) = 68.05

or

f (18) " 68 + ! .025(18 ! 20) = 68 + (!.025)(!2) = 68.05

16) C We are starting at the point (10,-8) and the function is increasing (because f’ is positive). We don’t know if the function will end above or below the x-axis, but since it is increasing, we are sure it must end somewhere above where it started at y = -8. Thus the only impossibility is ending up with a height of -10, which makes f ( 20) " !10 a true statement. All the other statements MAY be true.

17) D Since we are only given marginals (slopes), it is impossible to make a statement about whether there is a profit or loss, only whether the profit is increasing or decreasing. Using M" = MR ! MC we find

" ' (100) = 43 ! 40 = $3 / unit while " ' (101) = 44 ! 45 = !$1 / unit This book defines the change in profit on the nth item as M" (n ! 1) , so our estimate for the profit average rate of change from a production level of 100 to 101 units would be M! (100) . Since this is positive, the firm’s profit will increase. Our estimate for the profit average rate of change from a production level of 101 to 102 units would be M! (101) . Since this is negative, the firm’s profit will decrease, so only D is true. Note that whether profits increase or decrease not the same as a company making a profit or a loss. We do have to be careful about a similar statement for a general function as this definition only applies to cost, revenue, and profit. For a regular curve, it is possible for f ' (100) = 3 and f ' (101) = !1 yet the function from x=100 to x=101 to be decreasing. Ex.

18) E

d (10 " f ( x) ! 5 " g ( x) + 7) = 10 f ' ( x) ! 5 g ' ( x) + 0 . Plugging in x = 2 we get dx 10(18)-5(2)=180-10=170 19) B

d ( f ( x) ! g ( x)) = f ' ( x) g ( x) + g ' ( x) f ( x) (product rule) . Plugging in x = 0 we get dx (-18)*(1)+(2)*(-3) = -18-6=-24 20) D

d ( f ( g ( x)) = f ' ( g ( x)) ! g ' ( x) (chain rule) . Plugging in x = 1 we get dx

f ' ( g (1)) g ' (1) = f ' (3)(2) = 36(2) = 72